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.
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By inspection the solution is given by $$x=2$$ – Dr. Sonnhard Graubner Dec 25 '18 at 14:07
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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
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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.
KM101
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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.
Siong Thye Goh
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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