What to do with an exponential equation when there is a variable in the exponent of BOTH sides?

Question

Hi! Here's an example of what I mean:

3^x = 2^(x-1)

I took the log of each side, yielding:

log 3^x = log 2^(x-1)

OR:

x log 3 = (x-1) log 2

But what do I do from here?!?!?!? PLEASE HELP! I'll pick a best answer TODAY!

Answer 1

you're almost there....

x log 3 = x log 2 - log 2
x log 3 - x log 2 = - log 2
x (log 3 - log 2) = - log 2
x = (- log 2)/(log 3 - log 2)

REgards,
Chas.

Answer 2

You have done all the hard work already. Just move the x terms to one side and group the x as follows:

x log 3 - (x - 1) log 2 = 0
x log 3 - x log 2 + log 2 = 0
x ( log 3 - log 2) + log 2 = 0
x = -log 2 / ( log 3 - log 2)

Answer 3

x log 3 = x log 2 - log 2
x(log 3 -log 2) = - log 2
x * log(3/2) = - log 2
x = - log 2 / log (3/2)

Answer 4

3^x = 2^(x-1)

log(3^x) = log (2^(x-1))

x log 3 = (x-1) log 2
x log 3 = x log 2 - log 2

x log 3 - x log 2 = - log 2
x ( log 3 - log 2) = - log 2

x = - log(2) / (log(3) - log(2))

now log(3) - log(2) is log (3/2) or log(1.5)

so x = -log(2) / log (1.5)

x is roughly -1.7095

Answer 5

Looks good so far. Just solve for x:

x log3 = x log2 - log2 -->
x log 3 - x log2 = - log2 -->
x (log3 - log2) = -log2 -->
x= -log2/(log3-log2)

Therefore:

x = -1.71

Answer 6

(x-1)/x = log3/log2
1-1/x = log3/log2
1/x = 1 - log3/log2
x = 1/ (1 - log3/log2)