SOLUTION: I'm having trouble solving for 'x' in the equation {{{1.5(1.05)^x = 2(1.01)^x}}} I have tried turning it into a log, so I can solve for the approximate answer on a calculator, but

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I'm having trouble solving for 'x' in the equation {{{1.5(1.05)^x = 2(1.01)^x}}} I have tried turning it into a log, so I can solve for the approximate answer on a calculator, but      Log On


   

Question 98613This question is from textbook Calculus with Applications Brief Version
: I'm having trouble solving for 'x' in the equation 1.5%281.05%29%5Ex+=+2%281.01%29%5Ex I have tried turning it into a log, so I can solve for the approximate answer on a calculator, but I can't figure out how to get the 'x' out of the exponent on each side. Can you help me? This question is from textbook Calculus with Applications Brief Version

Found 2 solutions by Nate, stanbon:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
1.5%2A%281.05%29%5Ex+=+2%2A%281.01%29%5Ex
.
log%2810%2C1.5%2A%281.05%29%5Ex%29+=+log%2810%2C2%2A%281.01%29%5Ex%29
.
log%2810%2C1.5%2A%281.05%29%5Ex%29+-+log%2810%2C2%2A%281.01%29%5Ex%29+=+0
.

.

.
x%2Alog%2810%2C1.05%29+-+x%2Alog%2810%2C1.01%29+=+log%2810%2C2%2F1.5%29
.
x%28log%2810%2C1.05%29+-+log%2810%2C1.01%29%29+=+log%2810%2C4%2F3%29
.
x%2Alog%2810%2C1.05%2F1.01%29+=+log%2810%2C4%2F3%29
.
x%2Alog%2810%2C105%2F101%29+=+log%2810%2C4%2F3%29
.
x+=+log%2810%2C4%2F3%29%2Flog%2810%2C105%2F101%29

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1.5(1.05)^x = 2(1.01)^x
Divide both sides by 1.5 to get:
(1.05)^x = (4/3)(1.01)^x
Divide both sides by (1.01)^x to get:
[(1.05)/(1.01)]^x = (4/3)
[1.03960...]^x = (4/3)
Take the log of both sides to get:
xlog [......._] = log(4/3)
x = log(4/3)/log[....]
x = 0.120179....
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Cheers,
Stan H.