Solve

6^x+1 = 4^2x + 1

Printable View

- Nov 23rd 2008, 11:34 AMDanihopexponential powers
Solve

6^x+1 = 4^2x + 1 - Nov 23rd 2008, 11:43 AMTheEmptySet
start by subtracting 1 from both sides and taking the log

$\displaystyle 6^x=4^{2x} \implies \ln(6^x)=\ln(4^{2x}) \implies x\ln(6)=2x\ln(4)$

Collect your x's and factor

$\displaystyle x\ln(6)-2x\ln(4)=0 \implies x(ln(6)-2ln(4))=0$

Now by the zero factor theorem we have x=0.

Good luck - Nov 23rd 2008, 12:16 PMDanihop
I was trying to solve this problem a different way - I am at a loss - I don't know where to go from here - HELP PLEASE!

- Nov 23rd 2008, 12:36 PMo_O
What is wrong with TheEmptySet's response? Please try to be a bit more specific.

Also, I am curious ... Is your problem: $\displaystyle 6^{x+1} = 4^{2x+1}$ ... or ... $\displaystyle 6^x + 1 = 4^{2x} + 1$

TheEmptySet solved the second one for you. - Nov 23rd 2008, 12:50 PMDanihop
It is the first problem