Change of Base of Logarithm/Proof 2

Theorem

$\log_b x = \dfrac {\log_a x} {\log_a b}$

Proof

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We begin with an equation that relates two exponentials with two different bases, $a$ and $b$, both positive and neither equal to one:

$b^y = a^z$

Without loss of generality, select and solve for $b$:

$b = a^{z/y}$

Take logarithm's base $a$ of both sides:

\(\ds \log_a b\)

\(=\)

\(\ds \map {\log_a} {a^{z/y} }\)

\(\ds \)

\(=\)

\(\ds \dfrac z y \, \log_a a\)

Logarithm of Power

\(\ds \log_a b\)

\(=\)

\(\ds \dfrac z y\)

using $\log_a a = 1$

Now define $x = b^y = a^z$, and note that:

\(\ds \quad b^y = x\)

\(\implies\)

\(\ds y = \log_b x\)

Definition of General Logarithm

\(\ds \quad a^z = x\)

\(\implies\)

\(\ds z = \log_a x\)

Definition of General Logarithm

Substituting these values of $y$ and $z$ into our expression, $\log_a b = z / y$, yields the desired version of the change-of-base formula:

$\log_a b = \dfrac {\log_a x} {\log_b x} \implies \log_b x = \dfrac {\log_a x} {\log_a b}$

$\blacksquare$

Other choices involving $a$ and $b$

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This proof achieved the desired result only because two consecutive decisions were made regarding the selection of $a$ or $b$. Fortunately, there is little reason to remember these choices, provided the goal is to derive a useful formula. Suppose, for example, that we chose to solve for $b=a^{z/y}$, but instead take the logarithm of both sides at base $b$ instead of $a$. A slightly different change-of-base formula would be obtained:

$\log_b b=1=\log_b\left(a^{z/y}\right)= \dfrac{\log_a x}{\log_b x}\log_b a \implies \log_b x = (\log_b a)\cdot(\log_a x)$

Both formulas are useful, since:

$\left(\log_a b\right)\cdot\left(\log_b a\right)=1$

To understand why this identity is important, imagine that you are converting between base-2 and base-10, and have a numerical value for $\log_{10}2$, while your change-of-base formula involves $\log_2 10.$