Hi
If $\displaystyle \ln(\sqrt{x},\,10)$ is $\displaystyle \ln_{\sqrt{x}}10$ then simply use the definition of $\displaystyle \ln_ab=\frac{\ln b}{\ln a}$ :
$\displaystyle \ln_{\sqrt{x}}10=\frac{\ln 10}{\ln\sqrt{x}}=\frac{\ln 10}{\frac{1}{2}\ln x}=2\underbrace{\frac{\ln10}{\ln x}}_{\ln_x10}=2\ln_x10$
Hope that helps.
Hello, yuriythebest!
I constructed a general proof of this years ago . . .Show that: .$\displaystyle \log_{\sqrt{x}}(10) \;=\;2\cdot\log_x(10)$
$\displaystyle \text{Consider:}\:\log_b(N) \:=\:p \;\;{\color{blue}[1]}$
Then: .$\displaystyle b^p \:=\:N$
Square both sides: .$\displaystyle \left(b^p\right)^2 \:=\:N^2 \quad\Rightarrow\quad (b^2)^p \:=\:N^2$
Take logs (base $\displaystyle b^2$): .$\displaystyle \log_{b^2}(b^2)^p \:=\: \log_{b^2}(10^2)\quad\Rightarrow\quad p\cdot\log_{b^2}(b^2) \:=\:2\cdot\log_{b^2}(10^2)$
Since $\displaystyle \log_{b^2}(b^2) = 1$, we have: .$\displaystyle p \:=\:2\cdot\log_x(10)\;\;{\color{blue}[2]}$
Equate [1] and [2]: . $\displaystyle \boxed{\log_b(10) \;=\;2\cdot\log_{b^2}(10)}$
In baby-talk: "If we square the base, double the log."
And this can be generalized: "Cube the base, triple the log", etc.
.