The minimum value of $\log_{10}x+\log_x 10$

Question

Notation: $\log:=\log_{10}$

$\log x+\log_x 10$

$=\log x+ \frac{1}{\log x}$

$=\log(x \cdot \frac{1}{x})$

$=\log 1$

$=0$

Is the process correct? I doubt this is wrong. Please help. Thanks.

Answer 1

Note that $\log_a b=\frac{\ln b}{\ln a}$. Hence you want ot minimize $y+\frac1y$ with $y=\frac{\ln x}{\ln 10}$. "Clearly", this minimum is $2$ and achieved when $y=1$, i.e., when $\ln x=\ln 10$ and finally $x=10$.

Why $y=\frac1y\ge 2$ for $y>0$? Well, we have $y-2+\frac1y=\left(\sqrt y-\frac1{\sqrt y}\right)^2\ge 0$ with equality iff the expression in parentheses is zero.

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Your derivation is wrong when you go from $\log x+\frac1{\log x}$ to $\log(x\cdot \frac1x)$. Note that you'd need $\log x+\log \frac1{ x}$ for this instead of $\log x+\frac1{\log x}$.

Answer 2

I assume $\log_{10} x>0$ (iff $x>1$), because without constraints arbitrary small values can be gained.

$$\log_{10}x+\log_x 10=\log_{10}x+\frac{1}{\log_{10} x}\ge 2$$

with equality iff $x=10$, because $a+\frac{1}{a}\ge 2$ for $a>0$ with equality iff $a=1$.

To prove it, $$a+\frac{1}{a}\ge 2\iff \left(\sqrt{a}-\sqrt{\frac{1}{a}}\right)^2\ge 0$$ with equality iff $\sqrt{a}=\sqrt{\frac{1}{a}}\iff a=1$.