Find the value of the unknown variable in each of these:(i) 2^x = 128 , (ii) 3^y = 1/9 , (iii) 5^x = 625 , (iv) 4^s = 164 , (v) 16^t = 4 ,(vi) 8^ = 1/4 .- Solve each of the following equations for the value of x (i) log 3x = 6
ii) log[3]x + 3log[3](3x) = 3 iii) log (5x - 1) = 2 + log (x - 2) iv) 2logx = log(7x - 12) - 3. Write each of the following as a single logarithmic expression:
(i) log[10](x + 5) + 2 log[10] x (ii) log[4](x^3 - y^3) - log[4] (x - y)
iii) 1/2 (3 log[5] (4x) + log [5] (x + 3) - log[5] 9)
problem 2
ReplyDeleteii) log[3]x + 3log[3](3x) = 3
first bring all logs to one side of the equation.
check to see if the bases are similar
check for constants, if there is any, make them the power of the number that is being log.
remember that log a + log b = log a*b, and
log a - log b = log a/b
hence the equation can be rewritten as
log[3]x + log[3](3x)^3 = 3
log[3]x *(3x)^3 = 3
log[3]x * 9x^3 = 3
log[3] 9x^4 = 3
i will change the log function to base 10 by doing this
log 9x^4/ log 3 = 3
then cross multiply
log 9x^4 = 3 * log 3
the R.H.S can be worked out in the calculator since it is base 10
log 9x^4 = 1.431
9x^4 = log inverse 1.431
9x^4 = 27
x^4 = 27/9
x^4 = 3
x = 4√ 3
x = 1.32
i'm not sure if i'm correct though. if anyone has the right solution tell me
problem 2
ReplyDelete(i) log 3x = 6
i would find the inverse log of 6 then solve for x
3x = log inverse 6
x = 1000000/3
x = 333333.333
for problem 2 (ii) ghetto_celeb...u were correct until line log[3]x *(3x)^3 = 3....
ReplyDeletethe next line should have been..
log[3]x *27x^3 = 3..instead of log[3]x * 9x^3 = 3...because 3^3 is 27...
finishin the question log[3]x *27x^3 = 3
log[3] 27x^4=3
now using the rule :a^b=c...log[a] c=b
therefore from log[3] 27x^4=3
3^3=27x^4
27=27x^4
x^4=27/27
x^4=1
since 1 to any power equals 1...x=1
i don't understand wat u just said wattless bulb can u break it down a lil bit 4 me
ReplyDelete1. Find the value of the unknown variable in each of these:(i) 2^x = 128 , (ii) 3^y = 1/9 , (iii) 5^x = 625 , (iv) 4^s = 164 , (v) 16^t = 4 ,(vi) 8^ = 1/4 .
ReplyDeleteAns
Remember 2^3=8 this is log2^8=3
then to find the variable u log both 2 and 8 because as we all know our calculator only has log base 10
well i think that this is the same
i) 2^x=128 >> log2^128=x
x=log^128/log^2
x=7
ii)3^y = 1/9 >> log3^1/9=y
y=log^1/9/log^3
y=-2
iii) 5^x = 625 >> log5^625=x
x=log^625/log^5
x=4
iv) 4^s = 164 >> log4^164=s
s=log^164/log^4
s=3.7
v)16^t = 4 >> log16^4=t
t=log^4/log^16
t=0.5
vi) 8^ = 1/4
Can sum 1 help wit dis one i thinking the question has a typo but something tells me is jus a cleaver trick to get us going
no 3
ReplyDeletei) the 1st thing to do is to remove the coefficient:
log[10] (x+5) + log[10] x^2
now we know from our log rules that when adding common logs we multiply the log terms;
log[10] x^2(x+5)
and now we have one common log expression
no 3
ReplyDeleteii) we know from our log rules that when subtracting common logs we divide:
log[4] (x^3 - y^3) / (x-y)
now we have one log term
no 3
ReplyDeleteiii) the 1st thing to do is to remove the coefficients:
1/2 (log[5] 4x^3 + log[5](x+3) - log[5] 9)
BODMAS says that we should add 1st, but from our log rules we know that we have to multiply when adding:
1/2 (log[5] 4x^3(x+3) - log [5] 9)
now we have to subtract but we know from our log rules that this means we have to divide:
1/2 (log[5] (4x^3*x+3)/9)
now we have 1 log expression
thanks wattlessbulb for the help
ReplyDeletefor 1
ReplyDeletei) 2^x = 128
not sure if this is how it's supposed to be worked out but 128 is the same as 2^7 so x=7
3i) log[10](x + 5) + 2 log[10] x
ReplyDeletefrom laws of logs where log a + log b = log ab
log(x+5)(x)
log(x^2 + 5x)
1. Find the value of the unknown variable in i) 2^x = 128
ReplyDeleteI would use the very friendly equation:
2^3=8 = log2^8=3
where when 2^x = 128 is written in the log form, which is in the form showm on the righthand side of the above equation you will get log2^128=x , now you just punch the values in the calculator and you will get the answer for x. which is 38.53
i) 2^x=128 >> log2^128=x
x=log^128/log^2
x=7
ii)3^y = 1/9 >> log3^1/9=y
y=log^1/9/log^3
y=-2
iii) 5^x = 625 >> log5^625=x
x=log^625/log^5
x=4
iv) 4^s = 164 >> log4^164=s
s=log^164/log^4
s=3.7
v)16^t = 4 >> log16^4=t
t=log^4/log^16
t=0.5
vi) 8^ = 1/4
Can sum 1 help wit dis one i thinking the question has a typo but something tells me is jus a cleaver trick to get us going
major rule: 2^3=8 = log[2]^8=3
sory i copied tacobells answers to use as a guide. but i didnt get what they got, so i left the answers tacobell got so someone can correct me if im wrong , sory taco.
ReplyDeletewhat i did was up to the ans 38.53
ReplyDeletethe rest is tacobells
ReplyDeleteQUESTION 1.
ReplyDeleteAs miss teach us
2^3=8 >> log2^8=3
i) 2^x=128 >> log2^128=x we know to divide
x=log^128/log^2
x=7
ii) 2^3=8 >> log2^8=3
3^y = 1/9 >> log3^1/9=y We divide all
y=log^1/9/log^3
y=-2
iii) 2^3=8 >> log2^8=3
5^x = 625 >> log5^625=x
x=log^625/log^5
x=4
iv) 2^3=8 >> log2^8=3
4^s = 164 >> log4^164=s
s=log^164/log^4
s=3.7
v) 2^3=8 >> log2^8=3
16^t = 4 >> log16^4=t
t=log^4/log^16
t=0.5
vi) 2^3=8 >> log2^8=3
8^ = 1/4
log8^1/4=0 undefined
Plz correct me if i am wrong
Solve each of the following equations for the value of x (i) log 3x = 6
ReplyDeleteii) log[3]x + 3log[3](3x) = 3 iii) log (5x - 1) = 2 + log (x - 2) iv) 2logx = log(7x - 12)
i) log 3x = 6
rule = loga b = c is the same as a^c =b
a =10, b = 3x, c =6
10^6 = 3x
1000000 =3x
x = 1000000/3
x = 333333.333
ii) log[3]x + 3log[3](3x) = 3
rule = 3logb = logb^3
log[3]x + log[3](3x)^3 = 3
rule = lg[2]ab = lg2 a + log2 b
log[2] (x)*(3x)^3 = 3
log[2](x) * (27x^3) = 3
log[2] 27x^4 = 3
rule = loga b = c is the same as a^c =b
a = 2, b = 27x^4, c = 3
2^3 = 27x^4
8 = 27x^4
x^4 = 8/27
x^4 = 0.296
x = 4 root 0.296
x = 0.737
watlessbulb u lost me in problem 2(ii) wen u reach 27x^4.can u say hw u get dat..
ReplyDeletek ppl i seriously do not understand no.1.can sum1 explain it 2me????
ReplyDelete2^x = 128
ReplyDeleteapplying 2 ^ 3= 8 log [2] 8 = 3
2 ^ x = 128 log [2] 128 = x
x = 7
Number one is basically the first thing miss taught us in logs desi girl. Always remeber 2^3=8 is the same as log [2] 8 = 3. So in number one use this rule and you would solve the questions.
ReplyDeleteQuestion 1:
ReplyDeleteFrom the exponential form given, change to logs and the unknowns can be found..
2^x = 128
log 128/ log 2 = x
7 = x
3^y = 1/9
log(0.11)/ log 3 = y
-2 = y
5^x = 625
log 625/ log 5 = x
4 = x
4^s = 164
log 164/ log 4 = s
3.68 = s
16^t = 4
log 4/ log 16 = t
1/2 = t
8^x = 1/4
log 0.25/ log 8 = x
- 0.67 = x
Desi girl..sorry for the late reply...but how i got for 27x^4 problem 2(ii)....
ReplyDeletefrom the explanation...the line that says:
finishin the question log[3]x *27x^3 = 3
this means log (x multiply by 27x to the power of 3) base 3 =3
x multiply by 27x to the power of 3 is equal to 27x to the power of 4 ( because when multiplying powers they are added)
and 27x to the power of 4 is written as 27x^4
hope u understood that..kinda hard 2 explain here....
Yeh i agree seeing how the problem looks here would probly make you feel like you know nothing.
ReplyDelete2^x = 128
ReplyDeleteusing; 2^3 = 8 <=> log[2] 8 = 3
we get; log[2] 128 = x
log 128/log 2 = x
x = 7
the rest of no. 1 can be solved by the method that I used previously
ReplyDeletesorry but (vi) is missing the index, therefore, it can not be calculated!!!
ReplyDeleteiii) log (5x - 1) = 2 + log (x - 2)
ReplyDeleteplace all the logs on one side
then use the rule; log[a]b -log[a]c= log[a]b/c
then use=> 2^3 = 8 <=> log[2] 8 = 3 to change to exponential and solve
(i) log[10](x + 5) + 2 log[10] x
ReplyDeletelog is understood as log[10]
log (x + 5) + log x^2
when adding logs with the same base:
log [a]b + log[a]c = log[a](b*c)
we get=> log (x + 5) (x^2)
logx^3 + 5x^2
low rider i understand that concept of 2^3=8
ReplyDeleteso if i were to do ques.1 2^x = 128
it wood be log[2] 128=x
i understand up2 there..wat do i do frm here?????
#1:
ReplyDelete2^x=128
using 2^3=8 <=> (log8)/log2 =3
log128/log2 = x
therefore
x=7
#1:
ReplyDeleteii) 3^y = 1/9
using 2^3=8 <=> log2^8=3 (where 2 is the base no.)
(log1/9)/log3=y
#1
ReplyDeleteiii) 5^x = 625
using 2^3=8 <=> log2^8=3 (where 2 is the base no.)
log625/log5 = x
x = 4
(i) 2^x = 128
ReplyDelete2^3=8 <=> log2^8=3
always write this to avoid making errors
log 5 ^ 128= x
calculate dosent have base 5 so we need to ring it to base 10
log128/log5 = x
ii) log[3]x + 3log[3](3x) = 3
ReplyDeletei am having problems wih this question after you remove the coeff how do u solve for x???
(i) log[10](x + 5) + 2 log[10] x
ReplyDeletethis is already to the base 10 so we need to
remove the coeff
log10(x+5)+ log 10 ^2
i kind of stuck here could some one pleases help me??
well ken i think i can help here...
ReplyDeletelog[10](x + 5) + 2 log[10] x
since log[10] is same as log...we'l put 'log' to simplify it..so we get
log(x+5) + 2logx
get rid of coefficient so..
log(x+5) + logx^2
using rule --> loga + logb = log ab
soo.. log (x+5)(x^2)
so we get...
log (x^3 + 5x^2)