gledanju0

2022-06-26

Proving $\mathrm{log}{x}^{2}=2\mathrm{log}x$
How does $\mathrm{log}{x}^{2}=2\mathrm{log}x$
Can you do a proof please. I know that this is true but I don't know why.

scipionhi

Expert

Apply the exponential to get:
${e}^{\mathrm{log}\left({x}^{2}\right)}={x}^{2}={\left({e}^{\mathrm{log}\left(x\right)}\right)}^{2}={e}^{\mathrm{log}\left(x\right)}{e}^{\mathrm{log}\left(x\right)}={e}^{2\mathrm{log}\left(x\right)}$

Averi Mitchell

Expert

Assuming you have the following understaning of what is a logarithm:
$x={\mathrm{log}}_{b}a\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}a={b}^{x}$
So, let we have the following values:
$x={\mathrm{log}}_{b}a$ and $y={\mathrm{log}}_{b}c$
Which means that:
$a={b}^{x}$ and $y={\mathrm{log}}_{b}c$
If we multiply $a$ and $c$:
$a\cdot c={b}^{x}\cdot {b}^{y}={b}^{x+y}$
Hence, by our definition of logarithm:
$x+y={\mathrm{log}}_{b}\left(a\cdot c\right)={\mathrm{log}}_{b}a+{\mathrm{log}}_{b}c$ (do you remember that $x$ and $y$ are the logarithms of $a$ and $b$?)
So, we have prooved that:
${\mathrm{log}}_{b}\left(a\cdot c\right)={\mathrm{log}}_{b}a+{\mathrm{log}}_{b}c$
Now just use that formula in your particular case to get:
${\mathrm{log}}_{b}\left({x}^{2}\right)={\mathrm{log}}_{b}\left(x\cdot x\right)=\mathrm{log}\left(x\right)+\mathrm{log}\left(x\right)=2\mathrm{log}\left(x\right)$

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