SOLUTION: solve the equation {{{ 3^2x = 2^x-1.}}} round your answer to 3 decimal places please show me how to work this out because I am stuck also the exponent for 3 is 2x the exponen

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Question 1033654: solve the equation round your answer to 3 decimal places
please show me how to work this out because I am stuck
also the exponent for 3 is 2x
the exponent for 2 is x-1
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
solve the equation round your answer to 3 decimal places
please show me how to work this out because I am stuck
also the exponent for 3 is 2x
the exponent for 2 is x-1
================

2x*log(3) = (x-1)*log(2) = x*log(2) - log(2)
2x*log(3) - x*log(2) = -log(2)
x*(2log(3) - log(2)) = -log(2)
x = log(2)/(log(2 - log(9))
===============
x =~ -0.460845

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!

your equation is 3^(2x) = 2^(x-1)
take the log of both sides of the equation to get:
log(3^(2x)) = log(2^(x-1))
since log(a^b) = b*log(a), your expression becomes:
2x * log(3) = (x-1) * log(2)
simplify to get 2x * log(3) = x * log(2) - log(2)
subtract x * log(2) from both sides of the equation to get:
2x * log(3) - x * log(2) = - log(2)
factor out the x on the left side of the equation to get:
x * (2 * log(3) - log(2)) = - log(2)
divide both sides of this equation by (2 * log(3) - log(2)) to get:
x = - log(2) / (2 * log(3) - log(2))
simplify to get:
x = - .4608454206
that should be your answer.
replace x with that in the original equation of 3^(2x) = 2^(x-1)
you will get .3632801844 = .3632801844
that confirms the solution is correct.