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

Question

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.

Answer 1

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.