Logarithm Properties

Question

Does $$\log_{a^2}x=\log_a 2x$$

If it does, then how would you prove it?

If it doesn't, then how would I simplify $\log_{a^2} x$ so that it is in the form of $\log_a n$ where $n$ is some polynomial?

Answer 1

We have $$ \log{_{a^2}}{x}=\frac{\log x}{\log (a^2)}=\frac{\log x}{2\log a}=\frac12\log{_{a}}{x}=\log{_{a}}{(x^{1/2})}\ne \log{_{a}}(2x) $$ in general.

Answer 2

Recall the change of base formula, that says: $$\log_{b}a=\dfrac{\log a}{\log b}$$

So, $$\log_{a^2}x=\dfrac{\log x}{\log a^2}$$

$$\text {But we know, $\log a^2 = 2\log a$, from log properties}$$

$$\log_{a^2}x=\dfrac{\log x}{2\log a}$$

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Now, consider the other equation:

$$\log_{a}2x=\dfrac{\log 2x}{\log a}$$

This doesn't look equal..., you can find a counter example pretty quickly. Suppose $x=1$. We have: $\dfrac{\log x}{2\log a} = \dfrac{\log 1}{2\log a} = \dfrac{0}{2\log a} =0$

Other equation, we have: $\dfrac{\log 2x}{\log a} = \dfrac{\log 2*1}{\log a} = \dfrac{\log 2}{\log a}$

Clearly they are not equal.

Answer 3

Here's an explanation which doesn't explicitly use the change of base formula:

Let $c = \log_{a^{2}}(x)$

Then, from the definition of the $\log$ function:

$(a^{2})^{c} = x $

$\log(a^{2c}) =\log x$

$2c\log a = \log x \Rightarrow c = \frac{\log x}{2\log a}$

Let $d = log_{a}(2x)$

Then $a^{d} = 2x$

$d = \frac{2\log x}{\log a}$

$\Rightarrow d = 4c \Rightarrow d \neq c$