User:Guy vandegrift/sandbox

$\map {\log_a} {b^y}$ - Let:

Alternative proof
The advantages of this proof Express $x$ in terms of exponents of two different bases:
 * $b^y=a^z=x$

Solve for either base. Here we solve for $b$:
 * $b=a^{z/y}$

Take the logarithm of either base of both sides. First, we take the logarithm's base to be $b$:

Now use
 * $b^y = x \Rightarrow y = \log_b x$
 * $a^z = x \Rightarrow z = \log_a x$,

to obtain:
 * $\log a_b= \dfrac{\log_a x}{\log_b x}\Rightarrow \log_b x = \dfrac{\log_a x}{\log_a b}$

Aside: If we instead taking $b$ to be the base, we get a slightly different, but equivalent result:
 * $\log_b(b)=1=\log_b\left(a^{z/y}\right)= \dfrac{\log_a x}{\log_b x}\log_b a\Rightarrow \log_b x = \dfrac{\log_a x}{\log_b a}$

These results are equivalent because $\log_a b=\log_b a$.

Archives

 * Special:Permalink/551517 What is physics joke (failed to parse)
 * Special:Permalink/551563 Talk:Change of Base of Logarithm