# Math Help - Simple Log problem

1. ## Simple Log problem

Solve algebraically using logs.

$2^x = 5^x+1$

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

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?

2. Originally Posted by cantstop01
Solve algebraically using logs.

$2^x = 5^x+1$

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

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

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?
.

3. i am don't think it is right, but in this specific problem its true...
$2^x=5^x+1$
$2^x-5^x=1 /\times\log$
$\log(2^x)-\log(5^x)=\log(1)$
$x\times\log(2)-x\times\log(5)=\log(1)$
$x=\frac{0}{(\log(2)-\log(5))}$
$x=0$