How to Solve for x in...... 3^x-1=2^x Thank You!!!?

Answer 1

Take the log of both sides to get: (x-1)log3=xlog2
Solving this yields: x=log(3)/(log3-log(2))=log(3)/log(3/2)

Answer 2

i asked a question that is simular to this one.

my question was 3^x + 5^x = 10, which is similar to yours if you subtract 2^x and add 1 for both sides, which is 3^x - 2^x = 1, but this is the answer i got in this link below:

http://answers.yahoo.com/question/index;_ylt=AoLa.fazqmmTJ7xW4V0a2.Xty6IX?qid=20070209162619AAxloSO

but if you meant 3^(x-1) = 2^x, then i can solve it

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

(x-1) log3 = x log 2

xlog3 - log3 = xlog2

xlog3 - xlog2 = log3
x (log3/2) = log3
x = log3 / (log3/2)

Answer 3

3^(x - 1) = 2^x
(x - 1)ln(3) = xln(2)
xln(3) - ln(3) = xln(2)
xln(3) - xln(2) = ln(3)
x(ln(3) - ln(2)) = ln(3)
x = (ln(3))/(ln(3) - ln(2))