How does $\log^2(x)$ simplify to $\ln^2(x)/\ln^2(10)$?

Question

$\log^2(x) = \ln^2(x)/\ln^2(10)$

How does it simplify?

Answer 1

$\log^2(x)=(\log_{10}x)^2=\left(\dfrac{\log_ex}{\log_e10}\right)^2=\left(\dfrac{\ln x}{\ln10}\right)^2=\dfrac{\ln^2x}{\ln^210}$.

Answer 2

It's done using the Change-of-Base Formula for logarithms. Which is:

$ \log_b{x}=\frac{log_a{x}}{log_a{b}}$ being a the base where you want to change.

So as you have everything ^2 you can use the same formula but with all ^2 and this gives you:

$ \log_{10}{x}=\frac{log_e{x}}{log_e{10}}$ ==> $ (\log_{10}{x})^2=(\frac{log_e{x}}{log_e{10}})^2$

Answer 3

Let: $$\log_b(a) = \frac{\ln(a)}{\ln(b)}$$

Also

$$\log^2(x) = (\log(x))^2= \left( \frac{\ln(x)}{\ln(10)}\right)^2 = \frac{\ln^2(x)}{\ln^2(10)} $$

Note: $\ln(x) =\log_e(x)$ The Natural Logarithm

Answer 4

We want to prove $\log_ax=\frac{\log_bx}{\log_ba}$. Remember that by definition of logarithm we have $a^{\log_ax} = x$ for any base $a$ and hence we have:

\begin{align}\log_ax=\frac{\log_bx}{\log_ba}&\iff \log_bx = \log_ba\log_ax\\ &\iff b^{\log_bx} = b^{\log_ax\log_ba}\\ &\iff x = (b^{\log_ba})^{\log_ax}\\ &\iff x = a^{\log_ax}\end{align}

Now, $$\log^2 x = \left(\frac{\ln x}{\ln 10}\right)^2 = \frac{\ln^2x}{\ln^210}.$$