Logarithm Question

Solve $x + \log_{10} (1+2^x) = x\log_{10} 5 + \log_{10} 6$

Question

Logarithm Question

Solve $x + \log_{10} (1+2^x) = x\log_{10} 5 + \log_{10} 6$

mathslover · Answer

The first step I followed was :

$\log_{10} ({10}^x)  + \log_{10} ( 1 + 2^x) = x\log_{10} 5 + \log_{10} 6$

mathslover · Answer

Thus, LHS becomes : 

$\log_{10} ((10^x)(1+2^x)) = x\log_{10} 5 + \log_{10} 6$

mathslover · Answer

Thus, I get :

$(10^x)(1+2^x) = (5^x)(6) $

mathslover · Answer

$(2^x)(1+2^x) = 6$ 

$\implies 2^x + 2^{2x} = 6$  -------------- (1) 

Let $2^x = y$

Equation (1) becomes : 

$y + y^2 - 6 =0 $ or $y^2 + y - 6 =0 $

mathslover · Answer

Okay, so, using Quadratic Formula now : 

$y = \cfrac{-1 \pm \sqrt{1 + 24}}{2} = \cfrac{-1 \pm 5}{2} $ 

$\text{Thus y } \space = 2 $ or $-3$ 

Since, $2^x \ne -3$ 

So, $2^x = 2$ 

Thus, x = 1.

mathslover · Answer

o.O Made it more like a tutorial :P Yeah, well, I didn't know that I was on a right track but when I started writing that all, I just kept going... 

Though, I will appreciate your's suggestions and any short methods (if available).

kainui · Answer

Yeah, kinda cool how often it happens like this. If you see a problem that you know has an answer and you just start walking through it and believe in the math, then you often times just finish it without barely knowing haha

mathslover · Answer

Yeah, that's true! :)