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?

Printable View

- Apr 19th 2010, 01:19 AMcantstop01Simple 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 AMmathaddict
- Apr 19th 2010, 05:46 AMmathwhat
i am don't think it is right, but in this specific problem its true... (Thinking)

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

$\displaystyle 2^x-5^x=1 /\times\log$

$\displaystyle \log(2^x)-\log(5^x)=\log(1)$

$\displaystyle x\times\log(2)-x\times\log(5)=\log(1)$

$\displaystyle x=\frac{0}{(\log(2)-\log(5))}$

$\displaystyle x=0$