# Simple Log problem

• Apr 19th 2010, 01:19 AM
cantstop01
Simple Log problem
Solve algebraically using logs.

$\displaystyle 2^x = 5^x+1$

So far I've got $\displaystyle x log 2 = (x+1) log 5$

I think I'm going wrong on the next step. Am I to start isolating X by dividing log 2 by log 5 or is there something I'm missing?
• Apr 19th 2010, 02:09 AM
Quote:

Originally Posted by cantstop01
Solve algebraically using logs.

$\displaystyle 2^x = 5^x+1$

So far I've got $\displaystyle x log 2 = (x+1) log 5$

you only do this when the question is $\displaystyle 2^x=5^{x+1}$

I think I'm going wrong on the next step. Am I to start isolating X by dividing log 2 by log 5 or is there something I'm missing?

.
• Apr 19th 2010, 05:46 AM
mathwhat
i am don't think it is right, but in this specific problem its true... (Thinking)
$\displaystyle 2^x=5^x+1$
$\displaystyle 2^x-5^x=1 /\times\log$
$\displaystyle \log(2^x)-\log(5^x)=\log(1)$
$\displaystyle x\times\log(2)-x\times\log(5)=\log(1)$
$\displaystyle x=\frac{0}{(\log(2)-\log(5))}$
$\displaystyle x=0$