- 2(4 ^x-1) = 17^x
- log[2]P = log[2]23.9 + 0.45
- 2log x - 1 = log(1-2x)
- log[6]y = log[6]4-log[6]2
- log[3]y = 1/2 log[3] 7 + 1/2log[3]48
- (log[2]3)(log[2]y) - log[2]16 = 3
- log[5]x + log[5]30 = log[5]3 + 1
- 2(log[9] x + 2log[9]18) = 1
- 3log p = 2 + 3 log 4
- log[4]x + log[4]6 = log[4]12
- 2log[3]2 - log[3](x + 1) = log[3]5
- log[8] (x + 2) = 2 - log[8]2
- log (x+2) + log x = 0.4771
Wednesday, March 18, 2009
Logs set 1
Solve by explaining
Subscribe to:
Post Comments (Atom)
2(4 ^x-1) = 17^x
ReplyDeleteThe very first thing that I write is the equation relating logs to exponent.
2 ^ 3= 8 log base 2 8 = 3
I do not know how to proceed from here please help me.
log[2]P = log[2]23.9 + 0.45
ReplyDeleteI am assuming that []means base 2
common sense approach work out what can be worked out.
log[2] p = log [2] 24.35
because there is an = sign between the two and the base is common [2] you cancel it out
p = 24.35
correct me please anyone.
2log x - 1 = log(1-2x)
ReplyDeleteput in base 10 if it is not given
2 log [10] x-1 = log [10](1-2x)
remove all coefficents to powers.
log [10] x-1 ^2 = log [10]( 1-2x)
for question 2) Darky, you can't add 0.45 to log[2]23.9 which is what i believe you did.
ReplyDeleteinstead:
log[2]P = log[2]23.9 + 0.45
log[2]P = log[2]23.9 + 0.45log[2]2
since the log of any number to the same base is 1 (log[a]a = 1), we can perform the step above and use 0.45 as a coefficient to change the form of the number but not the value (0.45 x 1 = 0.45)
continuing:
since a{log(b)} = log(b^a)
log[2]P = log[2]23.9 + log[2](2^0.45)
log[2]P = log[2]23.9 + log[2](1.37)
then using log[a]b + log[a]c = log[a](b x c)
log[2]P = log[2](23.9 x 1.37)
log[2]P = log[2]32.65
now we can say antilog, or as you put it "cancel it out" becuase we have single logs on both sides of the equal sign with the same base.
therefore:
P = 32.65
i hope this clears things up for you. say so if it doesn't.
Darky
ReplyDelete2(4 ^x-1) = 17^x
What follows 2 ^ 3= 8 log base 2 8 = 3 is
isn;t the power that making the question lookhard
what is the power? log is the power
so take logs on both sides
log [2(4 ^x-1)] = log[17^x]
now log[17^x] will be x log 17
and log [2 X (4 ^x-1)] will be
log 2 + log (4 ^x-1)
and log (4 ^x-1) will be (x-1) log 4
can you take it from there
Darky
ReplyDeleteDJ halftime is correct you cannot add a log to a number
But darky you were correct in saying common sense work it out if possible
log[2]23.9 can be worked out
log 23.9 = 4.6
--------
log 2
so the question is now
log[2]P = 4.6 + 0.45
can you take it from there
Darky Darky
ReplyDelete2log x - 1 = log(1-2x)
put in base 10 if it is not given
2 log [10] x-1 = log [10](1-2x)
remove all coefficents to powers.
NOTE that the question is
2 log [10] x - 1 = log [10](1-2x)
so this would result in
log [10] x ^2 - 1 = log [10]( 1-2x)
what next group like terms together
log [10] x ^2 - log [10]( 1-2x) = 1
can you take it from there
Question 10
ReplyDeletesince log[4]x + log[4]6 = log[4]12
therefore log[4]x = log[4]12- log[4]6
rem log a -log b = log (a/b)
since all the base is the same, we divide their powers.
so x= 12/6
= 2
question 13
ReplyDeletelog (x+2) + log x = 0.4771
you will recognize log a + log b = log ab in the equation
rewriting to give
log (x+2)*x = 0.4771
expand to get rid of brackets on L.H.S
log x^2 + 2x = 0.4771
find log inverse of R.H.S to get rid of the log function
x^2 + 2x = log inverse of 0.4771
x^2 + 2x = 3
rewrite equation
x^2 + 2x - 3 = 0
use the quadratic formula to solve
(-b±√(b^2-4ac))/2a
a= 1, b=2, c=-3
x = (-2 ±√(2^2 - 4*1*(-3))/2*1
x = (-2 ±√(16)/2
x = (-2 ± 4)/2
x = 1 or x = -3
the log of a negative number will give an error in the calculator so you cannot use -3 for x
therefore:
x = 1
in question 10, you can also note that
ReplyDeletelog a + log b = log ab
so you can rewrite the question as
log[4]6x = log[4]12
since the bases are the same you can drop the logs to get
6x = 12
now you can solve for x.
x = 2
question 9
ReplyDelete3log p = 2 + 3 log 4
for this question i will bring all logs to one side
3 log p - 3 log 4 = 2
i would then get rid the constants infront of the log function by making it a power of the number after the log
log p^3 - log 4^3 = 2
i can see log a + log b = log ab from the equation
log p^3*4^3 = 2
log 12p^3 = 2
i'm not sure where to go from here
Some rules to remember in logs are:
ReplyDelete0) major rule: 2^3=8 = log[2]^8=3
1) log [7]^1=0 which simply means that once
the power is 1 it is equal to ZERO.
2) log[5]^5=1 which simply means that once the
base and the power is the same it is equal
to ONE.
3) log[2]^a + log[2]^b = log[2]^ab which
simply means when adding with the same base
you multiply the powers.
4) log[2]^a - log[2]^b = log[2]^a/b which
simply means that when minusing with the
same base you divide the powers.
5) b log[2]^a = log[2]^a^b which simply means
to remove coefficient which was b, this is
a very important thing to remember.
6) log b/a = log b/log a. log both the top and
bottom to obtain a value.
ALL OF THESE RULES ARE REVERSABLE.
SOMETHING TO REMEMBER.
- remove all coefficient
SOME METHODS USED TO SOLVE LOGS ARE;
- group all log terms
- making all log terms
- grouping all known terms
Thank you very much DJ HAlf time I say my mistake and understanded your approach very clearly
ReplyDeleteTaking from where miss said
ReplyDeletelog [10]x^2- log [10] 1-2x = 1
grouping like terms
log [10] ( x^2/1-2x) = 1
The x will cancel leaving a x
log [10] (x/1-2) = 1
log [10] x = 1
Because 1-2 = 1 and x / 1 =1
am I wrong
bye for now will be back very very soon I am lovin the blog i think that it is really helping me with maths
ReplyDeleteDarky i think u are almost right. there's a simple mistake. instead of X= 1,
ReplyDeleteX should b equal to -1.
since 1-2= -1 and x/-1= 1
therefore x= 1 (-1)
= -1
-a log is that which gets rid of powers
ReplyDelete-the inverse of the exponntial function is called the logerithmic function
2^3=8_______________ log[2]8=3
1. log[8] (x + 2) = 2 - log[8]2
ReplyDeletelog[8] (x+2) +log[8]2 =2
using the rule logA + logB = logA*B
log[8] (x+2)*2= 2
log[8] 4x +4 =2
4-2=log[8]4x
2=o.667x
X = 2/0.667
X =2.998
1. log[8] (x + 2) = 2 - log[8]2
ReplyDeletelog[8] (x+2) +log[8]2 =2
using the rule LOG A + LOG B = LOG AB
log[8] (x+2)*2= 2
log[8] 4x +4 =2
4-2=log[8]4x
2=o.667x
X = 2/0.667
X =2.998
correct me if iam wrong anyone
1. 2log[3]2 - log[3](x + 1) = log[3]5
ReplyDeleteusing the rule LOG A -LOG B = LOG A/B
2Log[3] 2/(x+1) = log[3]5
Log[3]2^2/(x+1) =log[3]5---->drop logs(same base)
4/(x+1)=5
4 = 5x + 5
5x =5-4
X = 1/5
X = 0.2
(2) Log 2 P = Log 2 23.9 + 0.45
ReplyDeleteThere are three solutions to this problem
Solution 1- Group all Log terms:
Log 2 P = Log 2 23.9 + 0.45
Log 2 [P/23.9] = 0.45
Using Log 2 8 =3 ↔ 2 3= 8
P/23.9 = 2 0.45
P/23.9 = 1.37
P = 1.37 * 23.9
P = 32.6
Solution 2- Bring al terms to logs:
Log 2 P = Log 2 23.9 + 0.45
Log 2 P = Log 2 23.9 + 0.45 Log 2 2_______using Log 2 2 = 1
Log 2 P = Log 2 23.9 + Log 2 2 0.45
Using Log x a + Log x b = Log x ab
Log 2 P = Log 2 [23.9 * 2 0.45 ]
Cancel logs: P = 23.9 * 2 0.45
P = 32.6
Solution 3- bring all Logs to the same base:
Log 2 P = Log 2 23.9 + 0.45
Using: Log a b = Log b/Log a
Log 2 P = [Log23.9/ Log2] + 0.45
Log 2 P = 5.03
Using Log 2 8 =3 ↔ 2 3= 8
2 5.03 = P
32.6 = P
please help me with this question
ReplyDelete2 Log x-1 = Log (1-2x)
Log (x-1) 2 = Log (1-2x)
Log x2 – 1 = Log (1-2x)
Bring all logs on one side:
Log x2 – Log (1-2x) = 1
Using: Log a b = Log b/Log a
Log [x2 /1-2x] = Log10_______since Log10=1
AM I GOING CORRECT? WHAT TO DO FROM HERE?
ok santosh i cant read de question too good but i have an answer so if u cud jus rewite it and i might be able to help u more is it log x minus 1 or X to th minus 1
ReplyDeleteHere is a question for someone to answer:
ReplyDelete2^2x – 2^x = 6
i find the above question an interesting log question to tackle
ReplyDelete9. 3log p = 2 + 3 log 4
ReplyDelete3logp -3log4 = 2
Logp^3 – log4^3 = 2
Logp^3 – log64 = 2
Log p^3/64 = 2
Log p^3 = 64*2
Log p^3 = log128
P^3 = 128
P =3√128
P = 5.039
can anyone correct me if i went wrong>????
Hey santosh... taking 2log x - 1 = log(1-2x) from the top:
ReplyDeletePut all logs on one side and nos by itself; (NB: the -1 in 2log x-1 is a number by itself), therefore:
2log x - log(1-2x) = 1
Also remember that if no base if given the base is 10.
Rewriting the equation with base [10] and also removing all coefficients:
log [10] x^2 - log [10] (1-2x) = 1
Recall: 1) log a - log b = log (a/b)
Therefore: log[10] (x^2/(1-2x) = 1
=> log[10] (x^2/(1-2x) = log [10]10
This is where u reached rite santosh. so lets continue;
since log on both sides of the equation has the same base it ca be canceled off.
=> x^2/(1-2x) = 10
from here you cross multiply:
=> 10 x (1-2x) = x^2
=> 10 - 20x = x^2
=> x^2 + 20x - 10
you'll get a quadratic equation.
So to solve for 'x' use the quadratic formula:
x = -b (+/-) √b^2 -4ac
-------------------------
2a
where: a=1, b= 20 and c = -10
when worked out you get two values of 'x':
I got: x = 0.5 and x = -19.5
I think that x=0.5 is the answer...
Can someone please confirm...
question 5
ReplyDeletefirst we look for the coefficients and remove them
so using the log rule
alog[b]c = log[b]c^a
log[3]y = log[3]7^1/2 + log[3]48^1/2
they all have the same bases
we can group the right side together
using the rule where
log[a]b + log[a]c = log[a](b*c)
so log[3]7^1/2 + log[3]48^1/2 = log[3]{(7^1/2)*(48^1/2)}
the numbers in the {} can be worked out
to get a whole number
we can not calculate the log of a number to the base of 3
so we use the log rule log[a]b = logb/loga
so by working out the right side of the = sign
we get a single number
let m represent our number
we get log[3]y = m
so using log rule that
log[a]b = c ~ a^c = b
so log[3]y = m ~ y = 3^m
that is an interesting question ghetto-celeb
ReplyDeletewhere 2^2x - 2^x = 6
first we recognize that 2^2x is the same as (2^x)^2
for example (2^3)*(2^3) is the same as 2^6 and
6=2*3
so we get (2^x)2 - 2^x = 6
we let 2^x = q
we get a quadratic equation
q^2-q=6
q^2 - q - 6 = 0
factorizing
we get
(q-3)(q+2)= 0
using 3
we get
2^x = 3
and (2^x)^2 = (3)^2 = 9
using any equation
we have
2^x = 3
so
x = log[2]3
or
2^2x = 9
so
2x = log[2]9
x = (log[2]9)/2
you all im not to good with this log thing but il try something.
ReplyDeletefor question 2:log[2]P = log[2]23.9 + 0.45
well in this case i will use the the rule
log b/a = log b/log a. where you log both the top and bottom to obtain a value.log b/a = log b/log a.
therefor you will get
log[2]P = log23.9/log[2] + 0.45
you now work out what you can which is the part log23.9/log[2] which gives a value of 4.58, now you place it back into the equation where you will get log[2]P = 4.58 + 0.45 , as you can see again all the known terms are grouped, where all values are on the right hand side and the logs are on the left hand side. now you can work out the value part where you will get log[2]P = 5.03
now using the rule 2^3=8 = log[2]^8=3 which is reversible you will get 2^5.03 = p
therefore p = 32.7
I THINK THATS CORRECT.
Question 7
ReplyDeletelet me c if i can do dis ques. caz am kinda lost rite nw....
log[5]x + log[5]30 = log[5]3 + 1
wel firstly the base is the same and wel wat i'l do nex is add the 3and 1.i don't no if am goin correct.so nw the ques. reads
log[5]x + log[5]30 = log[5]4
hmmm...then i'll brin the + log[5]30 on the rite side so it becomes negative but i not 2sure if 2leave it on the left side and work frm dey caz wen u minusin it becomes divide.k i'll go along wit d first sol'n.
log[5]x=log[5]4-log[5]30
so frm here i'l divide
log[5]x=log[5]4/30
since the base is the same ther's no need to rite it bac,frm here jus cancel out d logs so x=4/30
plz correct me if am wrong caz i not 2sure if it correct
o yea you can use the other methods i mentioned earlier to solve it as well but i find the gruoping of terms easier.
ReplyDeleteaye u no sumting ah goin try it d next way and c if ah go get d same ans.
ReplyDeletelog[5]x + log[5]30 = log[5]3 + 1
so as i said wat could work out work lik mad...k not 2mad but carefully..lol..
so it becom log[5]x + log[5]30 = log[5]4
k check dis log x/log5+log 30/log 5=log 4/log 5
k ppl dis d part u start workin out wat can work out,lik d log5,log30 and log4,they can al work out...so it nw becomes
logx/0.7+1.477/0.7=0.602/0.7
logx/0.7+2.11=0.86
logx/0.7=0.86-2.11
logx/0.7=-1.25
so frm here we cross multiply,so it becomes
logx=-1.25x0.7(dat is a multiplication sign)
logx=-.875
and since logs is only in base 10 it becomes
10^-.875=x (2^3=8 log[2]8=3)
0.133=x which iz d same ans. as 4/30
yeppppppy.....dam dat felt gud
logx
oh shocks i doh no way dat logx com frm in b buttom..jus ignore..wat u al are accustom doin..lol
ReplyDeletefor number 10, log[4]x + log[4]6 = log[4]12
ReplyDeletei will use the rule which states that when adding logs with the SAME base you must multiply the powers.
therefore you will get log[4]6x = log[4]12
i will now use the rule log b/a = log b/log a on the right hand side of the equation. where you log both the top and bottom to obtain a value.log b/a = log b/log a.
you will get log[4]6x = log12 /log4
you then work out what you can work out on the calculator which in this case is the righthand side. it will now be log[4]6x = 1.79
i will now use the rule 2^3=8 = log[2]^8=3
where you will get 4^1.79 = 6x which gives after woking 11.96 = 6x.
now you transpose and make x the subject of the formula which gives x = 11.69/6
x = 1.99
AND THAT IS HOW I WHOULD HAVE DONE IT AND I THINK IT CORRECT FROM MY INTERPRITATION.
wel desi girl taking from where you work it out like mad and got log[5]x + log[5]30 = log[5]4
ReplyDeletei would have used the same rule i jus used for question 10 which was when adding logs with the SAME base you must multiply the powers.
so taking the left hand side i will get
log[5]30x = log[5]4
then i will now use the rule log b/a = log b/log a on the right hand side of the equation. where you log both the top and bottom to obtain a value.log b/a = log b/log a.
you will get log[5]30x = log4/log5
you then work out what you can work out on the calculator which in this case is the righthand side. it will now be log[5]30x = 0.86
i will now use the rule 2^3=8 = log[2]^8=3
where you will get 5^0.86 = 30x which gives after woking 3.99 = 30x.
now you transpose and make x the subject of the formula which gives x = 3.99/30
x = 0.133
DJ half time i didn't no dat no.2 could b worked lik dat...caz i would hav worked mine lik dis...
ReplyDeletelog[2]P = log[2]23.9 + 0.45
wel i'll group d log terms 2gether jus lik u so it read lik dis
log[2]P-log[2]23.9=.45
and as d rule applies wen u minusin it becomes divide and wel since d base is d same then there no problem workin frm dey.wat i didn't no is if u could hav place log [2]2 in front of d .45.can u???
taking question 9, 3log p = 2 + 3 log 4
ReplyDeletewell firt of all you must get rid of the co-efficients.
you will then get log [p]^3 = 2 + log [4]^3
now using the rule log b/a = log b/log a on the right hand side of the equation. where you log both the top and bottom to obtain a value.log b/a = log b/log a.
you will now get log [p]^3 = 2 + log3/log [4]
now work out whatever you can which is on the right hand side where you will get
log [p]^3 = 2 + 0.79
log [p]^3 = 2.79
i will now use the rule 2^3=8 = log[2]^8=3
where you will get p^2.79 = 3 which gives
I DONT KNOW WHAT TO DO AGAIN SOMEBODY HELP!
k angel wel if i were u i'd say
ReplyDeletelogp^3=2+log4^3 jus as u said but instead i would hav worked out d 4^3 and get 64,so the equation nw reads
logp^3=2+log64
and jus work out log 64 which wel giv u 1.81 and add it to the 2
lik dis..logp^3=2+1.81
logp^3=3.81
since log is base 10 u it becomes 10^3.81=p^3
same 2^3=8 log[2]8=3
so 10^3.81=p^3
6456.5=p^3
18.62 =p
understand angel
no 7
ReplyDeletenotice that all terms have a common log
we know from the log rules that we multiply when adding common logs:
log^5 30x = log^5 4
since the = sign separates only 2 common log terms we can simply drop the log^5 and find for x:
30x = 4
therefore x = 0.13
hey guys for no 7 i made a mistake an used ^ for base instead of [] which is what miss uses. anyway im sure you know what i meant!!
ReplyDeleteno 8
in brackets, since we are adding common logs we know from our log rules that we can multiply. however we 1st have to apply another log rule in the brackets before multiplying. this is what i mean:
we have to remove the 2, so it will be:
2 log[9] 18 = log [9] 18^2
= log [9] 324
now we can multiply:
2(log[9] 324x) = 1
log[9] 324x = 1/2
now this can be put into exponent form:
9^1/2 = 324x
3 = 324x
x = 0.009
2 log x-1 = log (1-2x)
ReplyDeleteRemove coefficent to pwer and put in the base 10 if it is not given.
log [10] x-1^2 = log [10] 1-2x
apply log rule + or - and group like terms
log [1o] (x^2/1-2x) = 1
cancelling out the x
log[10] (x/1-2] = 1
log [10] x = 1
then applying log relating to exponent which says
2^3 = 8 log [2] 8 = 3
10^1 = x log [10]x = 1
x= 10
guys help me out please
no 9
ReplyDeletehey 'angel' i know you were getting a problem with no 9, and although 'Desi Girl' is correct, this is another way of solving the problem. i'll start from the beginning though:
you were correct when you said to get rid of the coefficients:
log p^3 = 2 + log 4^3
log p^3 = 2 + log 64
we can now work out everything on the right hand side:
log p^3 = 3.8
now we can simply use log inverse:
p^3 = log inverse 3.8
p^3 = 6309.57
p = 18.49
hope this method helps!!!
Question number 4
ReplyDeletelog [6] y = log [6] 4 - log [6] 2
log [6] y = log [6] (4/2)
log [6] y = log [6] 2
log [6] y = log 2/log 6
log [6] y = 0.387
applying logs relating to exponent
2^3 = 8 so log [2] 8 = 3
6 ^ 0.387 = y log [6] y = 0.387
y = 2.0
am i right or not ?
darky if you look futher up in the blog you will see that miss fariel helped you with that problem already!!!
ReplyDeleteI know that mysticwings I am takin it from where miss left out and am asking if i finished the question correctly
ReplyDeleteQuestion number 5
ReplyDeletelog[3] y = 1/2 log [3] 7 + 1/2 log [3] 48
I removed the coefficent to powers
log [3] y = log [3] 7 ^1/2 + 1/2 log [3] 48^1/2
common sense approach I worked out 7^1/2 and 48 ^1/2
log [3] y = log [3] 3.5 + log [3] 24
applying the log rule if i am right which says that if its addition you multiply
log [3] y = log [3] ( 3.5 * 24)
log [3] y = log [3] 84
common sense approach again I am goin to work out log base 3 * 84
log [3] y = log 84/log 3
log [3] y = 4.03
now using the log and exponent
2^ 3= 8 log [2] 8 = 3
3^4.03 = y log [ 3] y = 4.03
y = 83.7
correct of not ?
Miss Fariel can you plz answer this question:
ReplyDeletewhen you were correcting Darky for no 3, i was wondering why in the 9th line you squared x and 1 separately, ent you suppose to square it together and it will make a quadratic like this:
(x-1)^2 = x^2 - 2x + 1
can you plz explain it to me
Question number 6
ReplyDeletelog [3] * log [2] y - log [2] 16 = 3
CAN SOMOEONE HELP ME OUT AND EXPLAIN THE APPROACH
I think i should say
ReplyDeletelog [3] * log [2] (y * 16 )= 3
log [3] * log [2] 16y = 3
hey Darky question 4 is absolutely correct but you could of finished the problem a lot sooner. if you check your 3rd line you would realise that you have one log equal to another log with the same base, so all you had to do was to drop the log 6 on both sides and you would of gotten your answer : y = 2
ReplyDeletedo you understand???
hey Darky this is how to do no 6, by the way i think you wrote down the question wrong, anyway here is how to do it:
ReplyDeleteyou would notice that in no 6 everything in the left hand side has the same log base, but the first thing we have to do is to work out brackets of course:
we know from our log rules, when multiplying common logs we add:
log[2] (3+y) - log[2] 16 = 3
you would notice that the left hand side still has common logs and we know from our log rules that when subtracting common logs we divide:
log[2] (3+y)/16 = 3
now we can put this in exponent form:
2^3 = (3+y)/16
from her its just simple algebra:
8*16 = 3+y
128 = 3+y
y = 125
plz tell me if you understand!!!!!!!!!!1
Darky my darlin you are not working your calculator properly. for no 5 in your 4th line 7^1/2 is not equal to 3.5 it is 2.6 and similarly 48^1/2 = 6.9
ReplyDeleteafter this, you are very correct to apply the log rule and multiply, but why are you carrying on the problem so long? it can be solved striaght from the line you did your multiplication on. and as i told you before all you have to do at this point was drop your common log on each side and you would get your answer. let me show you:
log[3] y = log[3] (2.6 * 6.9)
log[3] y = log[3] 17.9
now dropping the log[3] we get:
y = 17.9
do you understand?
Here's one for someone to try...
ReplyDelete2 log[x] 2 + log[2] x = 3
#7) log[5]x + log[5]30 = log[5]3 + 1
ReplyDeleteThe first thing to do is check if all the base are the same.
Then recall: log a + log b = log ab
rewrite the eq:
log[5] 30x =log[5] 3 + 1
moving all logs on one side:
log[5] 30x - log[5] 3 = 1
recall: log a - log b = log (a/b)
So we get :
=> log[5] (30x/3) = 1
=> log[5] 10x = 1
remember: 2^3 = 8 <=> log[2] 8 = 3
5^1 = 10x
5 = 10x
x = 5/10
x = 1/2 =>ans
Here is a question
ReplyDelete2log[2]3-log[2]x=log[2]45
Question 11
ReplyDelete2log[3]2 - log[3](x + 1) = log[3]5
log[3]4-log[3]5=log[3](x+1)
Using log rule log a -log b =log a/b
log[3]4/5=log[3](x+1)
log[3]0.8=log3(x+1)
log0.8/log3 = log (x+1)/log 3
-0.20=log(x+1)/log3
-0.20log3=log(x+1)
-0.095=log(x+1)
-0.095=logX+log1
-0.095=logx
0.8=x
Is this correct not to sure tried another method.
low rider i think u were right up untilll....log[3]4/5=log[3](x+1)
ReplyDeletethe next line should have been..log[3]0.8=log[3](x+1)
then because the bases are the same([3])they can cancel leaving..0.8=x+1
therefore x=0.8-1
x=-0.2
and mysticwings* if u dont mind id like 2 ans the question u asked miss earlier....u squared the terms separately because the question was...
ReplyDelete2log x - 1 = log(1-2x)
not 2log (x - 1) = log(1-2x)
meaning that the entire term (2 log x) minus 1 is equal to log(1-2x)
so the coefficient 2 would become a power of x and not (x-1)
question2: log[2]P = log[2]23.9 + 0.45
ReplyDeletesince the two log terms are the same i belive they can cancle out with each other so you remain with:
p = 23.9 + 0.45
p = 23.45
question4: log[6]y = log[6]4-log[6]2
ReplyDeletesince logb(x/y) = logb x - logb y so :
log[6]y = log[6](4/2)
from here you are able to cancel out the log terms and you are left with :
y = (4/2)
y = 2
question5 : log[3]y = 1/2 log[3] 7 + 1/2log[3]48
ReplyDeleteto start solving this problem you will need to remove the figures in front the log therms and you are left with :
log[3]y = log[3] 7^o.5 + log[3]48^0.5
log[3]y = log[3] 2.646 + log[3]6.928
from here you can say that :
logb x + logb y = logb(xy)
so:
log[3]y = log[3] (2.646 x 6.928)
from here log terms are cancled out
y = 18.33
another way to aproch
ReplyDeletequestion5: log[3]y = 1/2 log[3] 7 + 1/2log[3]48
is:
since on one side there no unknown work out the logs by making sure there are no down stairs:
log[3]y = 1/2 (log7/log3) + 1/2 (log48/log3)
log[3]y = 1/2 (1.77) + 1/2 log(3.5)
log[3]y = 0.885 + 1.762
log[3]y = 2.647
from here you solve the simple equation
3^2.647 = y
18.32 = y
question6 : (log[2]3)(log[2]y) - log[2]16 = 3
ReplyDeletesince there is no unknowns in the first bracket it can be worked out :
(log3/log2) = 1.58
so :
1.58 log[2]y - log[2]16 = 3
remove the figure infront the log
log[2]y^1.58 - log[2]16 = 3
now since logb(x/y) = logb x - logb y :
log[2](y^1.58/16) = 3
so:
2^3 = y^1.58/16
8 = y^1.58/16
8x16 = y^1.58
128 = y^1.58
1.58^√ i28 = y
y = 21.56
question8: 2(log[9] x + 2log[9]18) = 1
ReplyDeletesince logb x + logb y = logb(xy)
then:
log[9](x^2 324) = 1
9^1 = x^2 324
9/324 = x^2
0.0277 = x^2
√0.0277 = x
x = 0.166
question9 : 3log p = 2 + 3 log 4
ReplyDeletefirst get rid of the figure in front the logs
log p^3 = 2 + log 4^3
since the base of the log isn't given it is understood that it is log[10]
log p^3 = 2 + log 64
log p^3 = 2 + 1.8
log p^3 = 3.8
from here you solve the simple equation keeping in mind that it is log[10]
10^3.8 = p^3
6309.5 = p^3
3^√6309.5 = p
p = 18.5
draky in your question you had 48 ^1/2 = 24 i think this is where u went wrong and trew off your answer it should be 48 ^1/2 = 6.928
ReplyDeletequestion10 : log[4]x + log[4]6 = log[4]12
ReplyDeletefirst we add the two log terms
log[4]x6 = log[4]12
from here we cancle out the logs on each sides since they are the same
x6 = 12
x = 12/6
x = 2
i have a question about one way on how to work number 5.
ReplyDeleteCan you calculate the log values on the right hand side since there are no variables then solve for y?
log[2]P = log[2]23.9 + 0.45
ReplyDeletefrom laws of logs...... log[a] B = X
X = (log B/log a)
(log P/log 2)= (log 23.9/log 2) + 45
(log P/0.3) = (1.38/0.3) + 45
(log P/0.3) = 9.1
log P = 2.73
P = 10^2.73
P = 537.03
@Blade, it seems possible to just calculate and get a value for the right side. it's a good idea to use, but personally i wouldn't for that question.
ReplyDeleteremember that your calculator can only calculate in base 10 or base e. so to get an actual figure for those logs you would have to do some changing of base to base 10. that would give you extra logs to work out making the problem just a bit tedious.
there are two attempts at question 5 above by fresh prince, both excellent. the second is what you're talking about Blade.
oh, and in his second approach, + 1/2 log(3.5) should be + 1/2(3.5), but other than that, good stuff man.
haha! yes Desi, you can do that in question 2. it's something i picked up from a friend. personally i like to use it because it makes th question look neat with all logs.
ReplyDeletewatch out though; if ever you use that approach make sure there isn't an easier way you think you can do it. i have created some problems for myself using that method when the problem could be worked much more easily.
but that's the beauty of maths, there's no fixed and rigid way to arrive at a solution. it's a very imaginative science.
The solution to getto celeb problem
ReplyDelete3log p = 2 + 3 log 4
wherre no base is given for the log the base is 10.
using the rule log b^3 = 3log b
log p^3 = 2 + log 4^3
which can be worken out using the calculator to simplify to give
log p^3 = 3.806
3logp = 3.806
log p = 3.806/3
log p = 1.268
Solution to Dark Angels problem
ReplyDelete2 log[x] 2 + log[2] x = 3
log[x] 2^2 + log[2] x = 3
log[x] 4 + log[2] x = 3
(log 4/ log x ) + (logx/ log2) = 3
0.6/logx + logx/0.3 = 3
now iam stick
Darky i think u dad ques. 3 a bit wrong caz u squared the 1 instead of the x.so it suppose to be, from where miss left of :
ReplyDeletelog[10]x^2-log[10](1-2x)=1
wel as one rule of logs applied,wen its a minus u divide.
so (log[10]x^2/log[1-2x=1
wel the logs wil cancel out and the square from the x wil wil cancel out wit the x from the 2in the denominator so u wood remain with
x/1-2=1 and wel darky u can find x frm there..
hey mysticwings i was wonderin the same ting, y miss squared the x and not the 1...can ne1 say y
ReplyDeleteppl for ques. 7 is it possible to do wat mysticwings did, add the x to the 30????
ReplyDeletesory mysticwings,i mean dark angel
ReplyDeleteQuestion 10
ReplyDeletelog[4]x + log[4]6 = log[4]12
log[4]x = log[4]12- log[4]6
log a -log b = log (a/b)
all the base is the same, so the powers can be divided.
therefore
x= 12/6
= 2
I agree with you desi girl i don't really know how dark angel just added x to 30 just so.
ReplyDeleteQuestion 10
ReplyDeletelog[4]x + log[4]6 = log[4]12
we carry across known on one side and unknown on the other side then we work out the right hand side first and its subtraction so we divide because the logs have the same base
log[4]x = log[4]12 - log[4]6
log[4]x = log[4] 12 / 6
log[4]x = log[4] 2
Now since the both side have the base logs we could remove the log
x = 2
hey Desi girl and low rider...
ReplyDeletedark angel did not just add 30 to x
dark angel used the rule...
(The first thing to do is check if all the base are the same.
Then recall: log a + log b = log ab)
when two logged terms with the same base are being added, u can multiply the two terms..hence
log a + log b = log ab
log[5]x + log[5]30 = log[5] 30x
and desi girl your question y miss squared the x and not the 1
ReplyDeleteu squared the terms separately because the question was...
2log x - 1 = log(1-2x)
not 2log (x - 1) = log(1-2x)
meaning that the entire term (2 log x) minus 1 is equal to log(1-2x)
so the coefficient 2 would become a power of x and not (x-1)...
leaving you with log x^2 - 1 = log(1-2x)
2(log[9] x + 2log[9]18) = 1
ReplyDeleteremove the coefficient of the 2nd term
2(log[9] x + log[9]18^2) = 1
2(log[9] x + log[9]324) = 1
removing the coefficient of the bracket,we get;
log[9]x + log[9 324) = 1/2
when adding logs, we simply multiply;
log[9](x * 324)= 1/2
changing to exponential form, we get;
9^1/2 =324x
3 = 324x
x = 3/324
x = 0.009
2(log[9] x + 2log[9]18) = 1
ReplyDeleteEXPAND THE EQUATION...
2log[9] x + 4log[9]18 = 1
REMOVE THE COEFFICIENTS
log[9] x^2 + log[9]18^4 = 1
WHEN ADDING LOGS WE MULTIPLY
log[9](X^2 * 18^4) = 1
CHANGE TO EXPONENTIAL FORM;
9 = X^2 * 18^4
ANYTHING TO THE POWER 1 IS THAT ANYTHING
9 = 104976X^2
9/104976 = X^2
X = 0.009
2log[3]2 - log[3](x + 1) = log[3]5
ReplyDeleteREMOVE THE COEFFICIENT
log[3]2^2 - log[3](x + 1) = log[3]5
WORK OUT THE INDICES
log[3]4 - log[3](x + 1) = log[3]5
APPLY RULE: LOG[a]b - LOG[a]c = LOG[a]b/c
HENCE: log[3]4/x + 1 = log[3]5
DIVIDE DY LOG[3]
4/x + 1 = 5
5(x + 1) = 4
5x + 5 = 4
5x = 4 - 5
x = -1/5
2log x - 1 = log(1-2x)
ReplyDeleteremove the coefficient
log (x - 1)^2 = log(1-2x)
you can divide by log 10 since log on L.H.S.=log on R.H.S.
(x - 1)^2 = 1 - 2x
expand the brackets
x^2 - x - x +1 = 1 - 2x
x^2 - 2x +1 = 1 - 2x
one side must be equal to zero
x^2 - 2x + 2x +1 - 1 = 0
x^2 = o
hence , x = 0
no. 3 was a bit tricky, could someone review it and reply
ReplyDeleteMr. Fresh Prince,WHAT IS THIS?
ReplyDeletep = 23.9 + 0.45
p = 23.45
well, let me correct you...
log[2]P = log[2]23.9 + 0.45
change the base of the log on the R.H.S.
remember: log[a]b = log[c]b/log[c]a
thus: log[2]P = 0.45 + log33.9/log2
note, when a base isn't seen, it is understood as base 10;
using ur calculator, divide the log
you'll get=> log[2]P = 0.45 + 4.58
adding, we get=> log[2]P = 5.03
changing toexponential, you'll get=>
P = 2^5.03
P = 32.7
here's another method
ReplyDeletelog[2]P = log[2]23.9 + 0.45
place all logs on one side
=> log[2]P - log[2]23.9 = 0.45
when subtracting logs, we divide
NOTE: log[a]b - log[a]c = log[a]b/c
hence: log[2]P/23.9 = 0.45
change to exponential format
=> P/23.9 = 2^0.45
P = 2^0.45 * 23.9
P = 32.6
k empress for no. 3 am not sure myself becaz i don't no if u can square the x-1 or jus the 1...can sum1 say plzz..
ReplyDeletek am empress if u look futher up miss had already helped darky with that ques.
ReplyDelete3log p = 2 + 3 log 4
ReplyDeletegroup the like terms
3log p - 3 log 4 = 2
remove the coefficientsof the logs
=> log p^3 -log 4^3 = 2
NB.log{10}x is understood
and log[a]b - log[a]c = log{a}b/c
thus: log p^3/4^3 = 2
log p^3/64 =2
changing to exponential form;
=>10^2 = P^3/64
10^2 * 64 = p^3
100 * 64 = p^3
p^3 = 6400
p = cube root of 6400
remember:
ReplyDelete2^3=8<==>log[2]8=3
log[7]1=0
7^0=1
log[5]5=1
5^1=5
log[2]ab=log[2]a+log[2]b
log[2]b/a=log[2]b-log[2]a
log[2]a^b=blog[2]^a
2. log[2] P = log[2] 23.9 + 0.45
ReplyDeletelog[2] P - log[2] 23.9 = 0.45
log[2] (P/23.9) = .45
2^.45 = p/23.9
23.9 (2^.45) = p
P = 32.6 mill
4. log[6]y = log[6]4-log[6]2
log[6]y = log[6] 4/2
log[6]y = log[6] 2
log[6]y = log[10]2 / log[10]6
log[6]y = .386
6^.386 = 1.99
5. log[3]y = ½ log[3] 7 + 1/2log[3] 48
log[3]y = log[3] 71/2 + log[3] 481/2
log[3]y = log[3]2.65 + log[3] 6.93
log[3]y = log[3] (2.65 × 6.93)
log[3]y = log[3] 18.36
log[3] 18.36 - log[3] Y
log[3] 18.36/4
log[3] 4.59
log[10] 4.59
log[10] 3
1.39
6. 2 (log[9] χ / 2 log[9] 18) = 1
(log[9] χ / log[9] 182 = ½
log[9] χ / log[9] 324 =1/2
log[9] χ/324 = ½
91/2 = χ/324
3 = χ/324
324 x 3 = χ
972 = χ/324
3log P = 2 + 3log 4
log P^3 = 2 + log 4^3
log P3 = 2 + log 64
log P3 - log 64 = 2
P3 = 2
64
64 x 2 = P3
128 = P3
5.1 = P
log [4]x / log[4]6 = log[4] 12
log[4] x/6 = log[4] 12
log x/6 = 1.79
4 1.79 = x/6
11.9 = x/6
71.71 = x
2log[3] 2 - log[3] (x +1) =log[3] 5
log[3] 22 - log[3] (x + 1) = log[3] 5
log[3] 4 - log[3] x +1 = log[3] 5
log (4/x + 1) = log 5 /log 3
log (4/x + 1) = 1.46
31.46 = 4/x + 1
4.97 = 4/x + 1
4.97 (4/x + 1) = 4
4.97χ + 4.97 = 4
4.97χ = 4 - 4.97
4.97χ = .97
log[2]P = log[2]23.9 + 0.45
ReplyDeletethe most important thing to remember is the rules
the first thing to write down is 2^3=8---> log2^8=3
log[2] P - log[2] 23.9 = 0.45
log[2] (P/23.9) = .45
2^.45 = p/23.9
23.9 (2^.45) = p
i having problems with this question could someone please help me solve it
ReplyDelete3ln2+in(x-1)=ln24
8) 2(log[9] x + 2log[9]18) = 1
ReplyDeletethe thing to remember when doin these question is to remove all coefficient
2(log[9] x + log[9]18^2) = 1
log[9]x + log[9 324) = 1/2
log[9](x * 324)= 1/2
change to exponential form
always remember to write this so u wouldn't make errors
2^3=8---> log2^8=3
9^1/2 =324x
3 = 324x
x = 3/324
x = 0.009
question 5 :
ReplyDeletelog[3]y = 1/2 log[3] 7 + 1/2log[3]48
focusing on the second part..
1/2 log[3] 7 + 1/2log[3]48
= 1/2 (log7/log3) + 1/2 (log48/log3)
= 1/2 (1.77) + 1/2 (3.52)
= 2.64
so,
log[3]y = 2.64
using rule 2^3 = 8 <--> log[2]8 = 3
we get y = 3^2.64
hence, y = 18.1