User:Guy vandegrift/sandbox

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$\map {\log_a} {b^y}$

Let:

Alternative proof

The advantage of this more tedious proof is that the symmetrical nature of the first equality between three entities is easier to remember, and ensures that a useful result is obtained, regardless of which base is selected in two crucial steps. This allows students to focus on the methods used in developing the desired formula, and not attempt to memorize the actual steps. We begin with an exponential expression involving two bases, $a$ and $b$:

$b^y=a^z=x$

The first step is to randomly select a base and solve for it. Here we solve for $b$:

$b=a^{z/y}$

Next, randomly select a base and take the logarithm of both sides to that base. Here we take $a$ as the base of the logarithm, but later show that a useful result occurs if the other base is chosen:

\(\ds \log_a b\) \(=\) \(\ds \log_a\left( a^{z/y}\right)\)
\(\ds \) \(=\) \(\ds \dfrac z y\, \log_a a\) Logarithms of Powers
\(\ds \) \(=\) \(\ds \dfrac z y\) Since $\log_a a=1$.

Now use

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

to obtain:

$\log a_b= \dfrac{\log_a x}{\log_b x}\Rightarrow \log_b x = \dfrac{\log_a x}{\log_a b}$

$\blacksquare$

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 $\log_a b=\log_b a$.

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