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.

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


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

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

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

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

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.

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

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