# exponential powers

• Nov 23rd 2008, 11:34 AM
Danihop
exponential powers
Solve

6^x+1 = 4^2x + 1
• Nov 23rd 2008, 11:43 AM
TheEmptySet
Quote:

Originally Posted by Danihop
Solve

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

start by subtracting 1 from both sides and taking the log

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

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

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

Good luck
• Nov 23rd 2008, 12:16 PM
Danihop
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 PM
o_O
What is wrong with TheEmptySet's response? Please try to be a bit more specific.

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

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