Solve the Following:
2 = 1.4^x
15 = 6.1^x-1
6 = 10^x+1
14^x = 6^x
4 = 3.3^x
18 = 3.1^x
Find b in the following:
log[4]16 = b
log[3]81 = b
log[2]16 = b
log[5]125 = b
log[3]9 = b
log[6]216 = b
log[8]64 = b
log[4]64 = b
log[b]625 = 4
log[b]49 = 2
log[b]27 = 3
log[b]81 = 2
log[b]4 = 2
log[b]16 = 4
log[3]b = 6
log[4]b = 3
log[5]b = 3125
log[10]b = 1
log[9]b = 4
log[2]b = 6
log[3]b = 4
log[8]b = 3
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2 = 1.4^x
ReplyDeletelog1.4 2=x
since diff base u divide it
log1.4/ log 2 = x
0.49= x
log[4]16 = b
log4/log 16= b
0.5= b
log[3]b=4
ReplyDeletelog[2]8=3--> 2^3=8
therefore
3^4=b
hence b= 81
log[2]16=b
ReplyDeletechanging to indices
since log[2]8=3--> 2^3=8
2^b=16
therefore
2^4=16
hence b=16
log[b]49=2
ReplyDeletesince
log[2]3=8--> 2^3=8
then
b^2=49
hence
b=7
2 = 1.4^x
ReplyDeleteis not 1.4^x = 2 the same as the question
thus 2^3=8 log2^8=3
log1.4^2 = x
x = 2.06
log[5]125...
ReplyDeletethis problem can be solved by following the rule *log[m]v = logv / logm
therefore :
log125 / log 5 = 3
18= 3.1^x
ReplyDeletein this case we can use the approach following 2^3 =8...so log[2]8 = 3
therefore we get :
3.1^x = 18 ....log[3.1]18 = x
from here we use the rule log[m]v = logv / logm
hence:
log18 / log3.1 = x
x=2.55 or 2.6
log[3]b = 6
ReplyDeleteso again following 2^3 =8... log[2]8 = 3
we get:
log[3]b = 6....log3^6 = b
hence :
b= 729
remember this step can be comfusing at times so like miss said we can write the rule 2^3 =8 over the question in pencil or something and the answer tends to become clearer!
log[b]625 = 4
ReplyDeletei remember 2^3 = 8 is log[2] 8 = 3
i remember to convert to log base 10 work out with my calculator
so log 625/ log b = 4
cross multiply
log 625/ 4 = log b
work out L.H.S with calculator
0.699 = log b
find log inverse of L.H.S to solve for b
log invs 0.699 = b
5 = b
18 = 3.1^x
ReplyDeletei remember 2^3 = 8 is the same as log[2] 8 = 3
so log[3.1] 18 = x
i will change the L.H.S to base 10 to work it out in the calculator
log 18/ log 3.1 = x
2.55 = x
log[b] 27 = 3
ReplyDeletewe can use one of the log rules: 2^3 = 8 this is log[2]8 = 3
b^3 = 27
b = cube root of 27
b = 3
log[4] b = 3
ReplyDeleteput this in exponent form and solve:
4^3 = b
b = 64
log[4] b = 3
ReplyDeletethis can be worked out exactly as the one above because its the same problem but different values are used:
b = 5^3125
log[10] b = 1
ReplyDeletethis can be put in exponent form:
10^1 = b
b = 10
Hey there!
ReplyDeleteQuestion: log[4]64 = b
Answer:
recall; 2^3 = 8, which implies log[2]8 = 3
Now, for log[4]64 = b we consider what MOM told us;
Bring logs to base 10 by drawing a line to separate the log[4]64 into:
log64/log4 = b
therefore, b = log64/log4
2=1.4^x
ReplyDeleteyou state that 2^3=8 then log[2]8=3
the 2 is now 1.4
the 3 is now x
the 8 is now 2
so then you can find x, log2/log1.4
0.301/0.146
x=2.1
4=3.3^x
ReplyDelete2=3.3
3=x
8=4
log4/log3.3
0.0602/0.518
x=1.2
18=3.1^x
ReplyDelete2=3.1
3=x
8=18
log18/log3.1
1.255/0.491
x=2.55
log[4]16=b
ReplyDeletelog16/log4
1.204/0.602
b=2
Question
ReplyDeletelog[2]b=6
As a key point to recall as MOM states to remember:2^3 =8...so log[2]8 = 3
therefore the next step is to
log[2]b=6 ; 2^6=b
b=64
From
ReplyDeletelog[3]b = 4
Key to answer any question
2^3 =8...so log[2]8 = 3
Then,log[3]b = 4 ; 3^4=b
b=81
2 = 1.4^x
ReplyDeletelog1.4 2 = x
From the laws of logs log a (B) = X
(log B/log a)=x
(log2/log1.4)= x
x = (0.30/0.15)
= 2
PART 2
ReplyDeletea. log[4]16 = b
we expand the values multipling them by log
log16/log4
1.204/0.602
b=2
b. log[3]81 = b
log81/log3
b = 4
c. log[5]125 = b
log125/log5
b = 3
d. log[3]9 = b
log9/log3
b = 2
e. log[6]216 = b
log216/log6
b = 3
f. log[8]64 = b
log64/log8
b = 2
g. log[4]64 = b
log64/log4
b = 3
log[3]81=b
ReplyDeletelog81/log3
1.908/0.477
b=4
log[2]16=b
log16/log2
1.024/0.301
b=4
log[5]125=b
log125/log5
2.096/0.698
b=3
log[3]9
log9/log3
0.954/0.477
b=2
log[8]64
log64/log8
1.806/0.903
b=2
log[6]216
log216/log6
2.334/0.778
b=3
log[4]64
log64/log4
10806/0.662
b=3
2 = 1.4^x
ReplyDeletelog1.4 2=x
since the log is not at base ten, simply divide the two logs, which is as follows:
log1.4/ log 2 = x
0.49= x
2)log[4]16 = b
log4/log 16= b
0.5= b
Alwas remember the following information when attempting any log question
ReplyDelete2^3 =8
therefore log[2]8 = 3.
this information will help in solving logs question
log[b] 27 = 3
ReplyDeleteSwitch to exponent by using the rule in the previous comment above.
Therefore;
b^3 = 27
b = cube root of 27
b = 3
log[4] b = 3
Switch to exponent by using the rule in the previous comment above.
4^3 = b
b = 4 * 4 * 4
b = 64
log[b]625=4
ReplyDeletewell i'm not sure how to do this quet. but i will try
log625/logb=4
now you log both sides
4logb=log625
4logb=2.795 now i'm stuck and i don't know what to do can someone help me please
RTS:
ReplyDelete2=1.4^x
STEP1: TAKE LOG on both sides
log 2 = log 1.4 ^x
log 2 = x log 1.4
step2: group like terms
x =log 2 / log 1.4
x =2.06
rts:
ReplyDelete15=6.1^x-1
step 1:take log on both sides
log 15 =log 6.1^ x-1
log 15 =(x-1) log 6.1
step 2: group like terms
x-1 = log15 / log 6.1
x-1 = 1.50
x =1.5+1
x =2.5
Can someone help me with this problem
ReplyDelete14^x = 6^x
I have dont exactly know where to begin
log[4]^16=b
ReplyDeletewell looking at what some of them did
I would use the major rule of
log[2]^8=3 = 2^3=8
thus 4^b=16
therefore to obtain a value to be equal to 16
i would use 2 as the power for 4 as the base cause 4 squared is 16.
4^2=16
therefore b=2
i would basically do the same things for the next 7 questions as they are all in the same format.
ReplyDeletefor the questions with the base = b for example log[b]625 = 4
ReplyDeleteI would use the major rule of
log[2]^8=3 = 2^3=8
thus b^4=625
therefor to obtain a value with a power of 4 to be equal to 625. the base will be 5 as 5^4=625
therefore b=5.
for the questions with the power = b for example log[3]b = 6
ReplyDeletei would also use the major rule of
log[2]^8=3 = 2^3=8
thus 3^6=b
now you just punch it into the calculator and you get b=729
This comment has been removed by the author.
ReplyDeletelog[3]81=b
ReplyDeletelog81/log3
1.908/0.477
b=4
log[2]16=b
log16/log2
1.024/0.301
b=4
log[5]125=b
log125/log5
2.096/0.698
b=3
log[3]9
log9/log3
0.954/0.477
b=2
log[8]64
log64/log8
1.806/0.903
b=2
log[6]216
log216/log6
2.334/0.778
b=3
log[4]64
log64/log4
10806/0.662
b=3
2 = 1.4^x
ReplyDelete2 ^3= 8 log 2 8=3
1.4^x = 2 log 1.4 2 = x
log2/log1.4 =2.06
for the question "swanky thing" asked on
ReplyDelete14^x = 6^x
you can log both sides and get
log14^x = log6^x =
from the rule the powers can be placed at the front=
xlog14 = xlog6
now this question probably has an error because the variable would all cancel as you can see xlog can cancel the other xlog because they have the same base and an equal sign exists.
after canceling there would only be 14 = 6 and this to me isnt logical. well have to ask miss about this question just to make sure.
I see where your comming from in the logics of it "etioncen" but i really have to check this out with miss.....
ReplyDeleteYes i agree this question probly has an error,but could we make x subject of formula and solve instead of cancelling out log.
ReplyDelete4 = 3.3^x
ReplyDeleteapplying 2 ^3 =8
log [2] 8 = 3
3.3 ^x = 4
log [3.3] 4 = x
1.16 = x
6 = 10^x+1
ReplyDeleteapplying 2^ 3= 8 log [2] 8 = 3
10^ x + 1= 6 log [10] 6 = x + 1
0.7782 - 1 = x
x = -0.222
a) log[4]16 = b
ReplyDeletelog16/log4
1.204/0.602
therefore b=2
b) log[3]81 = b
log81/log3
therefore b = 4
c) log[5]125 = b
log125/log5
therefore b = 3
d) log[3]9 = b
log9/log3
therefore b = 2
e) log[6]216 = b
log216/log6
therefor b = 3
finding b.......
ReplyDeletea)log(4)16=b
log16/log4=b
1.204/0.602=b
therefore
b=2
b)log(3)81=b
log81/log3=b
1.908/.477=b
b=4
c) log(2)16=b
log16/log2=b
1.204/0.301=b
b=4
d) log(5)125=b
log125/log5=b
2.097/0.699=b
b=3
e) log(3)9=b
log9/log3=b
.954/.477
b=2
f) log(6)216=b
log216/log6=b
2.33/.778=b
=2.99
b=3
g) log(8)64=b
log64/log8=b
1.806/0.903
=2
h) log(4)64=b
log64/log4=b
1.806/0.602
=3
hey miss i really bad at logs but as i read some of these postings and this is wat i get,
ReplyDeletelog (3) 81 = b
since log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 3^b = 81
hence 3^4 = 81
that means b= 4
log (2) 16 = b
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 2^b= 16
hence 2^4= 16
that means b= 4
log (5) 125= b
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 5^b= 125
hence 5^3= 125
that means b=3
log (3) 9= b
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 3^b= 9
hence 3^2=9
that means b=2
log (6) 216=b
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 6^b=216
hence 6^3=216
that means b=3
log (8) 64=b
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 8^b= 64
hence8^2=64
that means b=2
log (4) 64=b
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 4^b=64
hence 4^3=64
that means b=3
log (b) 625= 4
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore b^4=625
hence 5^4=625
that means b=5
log (b) 49 = 2
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore b^2=49
hence 7^2=49
that means b=7
log (b) 27 = 3
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore b^3=27
hence 3^3=27
that means b=3
log (8) b = 3
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 8^3=b
that means b=512
log (8) b = 3
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 8^3=b
that means b=512
log (3) b = 4
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 3^4=b
that means b=81
log (2) b = 6
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 2^6=b
that means b=64
log (9) b = 4
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 9^4=b
that means b=6561
log (10) b = 1
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 10^1=b
that means b=10
log (5) b = 3125
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 5^3125=b
that means b= really really big
wel i think that these correct, any1 thinks there wrong
Part 2:
ReplyDeleteFor all the below questions, since the calculator only has log base 10. simply divide the both values using the log base 10. (this can be also considered as expansion)
log[4]16 = b
log 16/ log 4 = b
2 = b
log[3]81 = b
log 81/ log 3 = b
4 = b
log[2]16 = b
log 16/ log 2 = b
4 = b
log[5]125 = b
log 125/ log 5 = b
3 = b
log[3]9 = b
log 9/ log 3 = b
2 = b
log[6]216 = b
log 216/ log 6 = b
3 = b
log[8]64 = b
log 64/ log 8 = b
2 = b
log[4]64 = b
log 64/ log 4 = b
3 = b
Part 2
ReplyDeleteFor all the below question, the logs can be switched to the exponent form in order to find for b...
log[3]b = 6
3^6 = b
729 = b
log[4]b = 3
4^3 = b
64 = b
log[5]b = 3
5^3 = b
125 = b
log[10]b = 1
10^1 = b
10 = b
log[9]b = 4
9^4 = b
6561 = b
log[2]b = 6
2^6 = b
64 = b
log[3]b = 4
3^4 = b
81 = b
log[8]b = 3
8^3 = b
512 = b
Part 2
ReplyDeleteAll the below questions can be switched to exponent type to find the value of b:
log[b]625 = 3
b^3 = 625
b = cube root 625
b = 8.55
log[b]49 = 2
b^2 = 49
b = square root 49
b = 7
log[b]27 = 3
b^3 = 27
b = cube root 27
b = 3
log[b]81 = 2
b^2 = 81
b = square root 81
b = 9
log[b]4 = 2
b^2 = 4
b = square root 4
b = 2
just read some posts and so far i agree with mostly everyone
ReplyDeletefor the first set:
i would use the strategy where 2^3=8 <=> log2^8(where 2 is the base of log)
using the first question as an example for my above post:
ReplyDelete1.4^x=2
therefore
log2/log1.4=x
2.06=x
that formulae could be used for all of the questions in the first set.
ReplyDeleteread some more comments an it seemed mostly everyone decided to do it the same way
for the second set of questions the formulae i would use is:
ReplyDeletelog[m]v= logv/logm
this, i find is the easier approach to the question
as an example to the above post:
ReplyDeletelog[4]16=b
using log[m]v= logv/logm
log16/log4 = 2
log[8]b = 3
ReplyDeleteStart with saying 2^3=8 log [2] 8= 3
so log [8]b = 3
8^3 = b
b =512
log[2]16=b
ReplyDeletechanging to indices
since log[2]8=3--> 2^3=8
2^b=16
therefore
2^4=16
hence b=16
log (4) 64=b
ReplyDeletesince log 2 ^ 8 = 3 2 ^ 3 = 8
therefore 4^b=64
hence 4^3=64
that means b=3
2 = 1.4^x
ReplyDeletelog1.4 2=x
log1.4/ log 2 = x
0.49= x
log[4]16 = b
log4/log 16= b
0.5= b
log[4] b = 3
ReplyDeleteput this in exponent form and solve:
4^3 = b
b = 64
log[b]49=2
ReplyDeletesince
log[2]3=8--> 2^3=8
then
b^2=49
hence
b=7
log[3]b = 4
ReplyDelete3^4 = b
81 = b
log[b]625 = 3
ReplyDeleteb^3 = 625
b = cube root 625
b = 8.55
log[b]27 = 3
ReplyDeleteb^3 = 27
b = cube root 27
b = 3
log[b]81 = 2
ReplyDeleteb^2 = 81
b = square root 81
b = 9
4 = 3.3^x
ReplyDeleteusing the rule 2^3 = 8 <--> log[2]8 = 3
we get, x = log[3.3]4
rule..log mv = log m/log v
we get, x = log4 / log 3.3
so .. x = 1.16
log[10] b = 1
ReplyDeleteusing rule.. 2^3 = 8 <--> log[2]8 = 3
we get , 10^1 = b
hence b = 10
another approach to the question above...
ReplyDeleteif we remember the other rule that...
log of any number to the same base = 1
eg. log[3] 3 = 1
then we can just say
log[10] b = 1
hence, b = 10
log[3] b= 6
ReplyDeleteusing rule.. 2^3 = 8 <--> log[2]8 = 3
we get, b = 3^6
so, b = 729
another apporoach to the question above is :
ReplyDeletelog[3] b= 6
using rule --> log mv = log m/log v
we get, log b / log 3 = 6
log b = 6 (log3)
log b = 2.86
so, b = log^-1 (inverse) 2.86
hence, b = 729