1. what rule?

can you please explain how this occurs:

2. 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.

3. Originally Posted by yuriythebest
can you please explain how this occurs:

What machine and what application is this from?

RonL

4. Hello, yuriythebest!

Show that: .$\displaystyle \log_{\sqrt{x}}(10) \;=\;2\cdot\log_x(10)$
I constructed a general proof of this years ago . . .

$\displaystyle \text{Consider:}\:\log_b(N) \:=\:p \;\;{\color{blue}[1]}$

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.
.

5. thanks everyone!

captainBlack- this was from the microsoft student graphing calculator. very user friendly.