# Solve $\log_{10}{(10^x+100)} = \frac{1}{2}x+1+\log_{10}2$

I know that $$\frac{1}{2}x+1+\log_{10}2$$ can be manipulated to become $$\log_{10}{10^{\frac{1}{2}x}}+\log_{10}10+\log_{10}2$$ and $$\log_{10}20*10^{\frac{1}{2}x}$$, but I don't see how $$\log_{10}{(10^x+100)} = \log_{10}20*10^{\frac{1}{2}x}$$ can be solved.

• By inspection the solution is given by $$x=2$$ – Dr. Sonnhard Graubner Dec 25 '18 at 14:07
• Well why didn't I think of that. :P I am however curious about the algebraic solution. – Nameless King Dec 25 '18 at 14:15

Hint:

$$\log_{10}{(10^x+100)} = \frac{1}{2}x+1+\log_{10}2$$

$$\log_{10}{(10^x+100)} = \log_{10}(20\cdot 10^{\frac{1}{2}x})$$

$$10^x+100 = 20\cdot 10^{\frac{1}{2}x}$$

$$(10^{\frac{1}{2}x})^2+100 = 20\cdot 10^{\frac{1}{2} x}$$

Here, let $$t = 10^{\frac{1}{2} x}$$ to reach a quadratic equation.

Guide:$$\log_{10}\left(\frac{10^x}2 +50\right)=\frac12x+1$$

$$\frac{10^x}2 +50=10^\left(\frac12x+1\right)$$

Let $$y = 10^\frac{x}2$$ and solve a quadratic equation.

• and where is $$\log_{10} 2$$? 0k, it is in the left-hand side of the equation. – Dr. Sonnhard Graubner Dec 25 '18 at 14:09