TAOCP 1.2.2 Exercise 14

We are asked to prove Eq.

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Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 14. [15] Prove Eq. (12).

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Solve time: 44s

Solution

We are asked to prove Eq. (12), which states:

$\log_b(c^y) = y , \log_b c, \quad \text{if } c > 0. \eqno(12)$

Let $b > 0$ and $b \ne 1$ be fixed. By definition, the logarithm $\log_b z$ of a positive real number $z$ is the unique real number $x$ such that

$b^x = z. \eqno(9)$

Set $z = c^y$ with $c > 0$ and $y$ real. Let

$x = \log_b(c^y).$

Then, by definition of logarithm,

$b^x = c^y. \eqno(20)$

Consider $u = \log_b c$. By definition,

$b^u = c. \eqno(21)$

Raising both sides of (21) to the power $y$ and applying the law of exponents (5) for real numbers, we obtain

$b^{uy} = (b^u)^y = c^y. \eqno(22)$

Comparing (20) and (22), we see that

$b^x = b^{uy}.$

Since $b > 0$ and $b \ne 1$, the exponential function $b^t$ is strictly increasing and hence injective on $\mathbb{R}$. Therefore, equality of the powers implies equality of the exponents:

$x = uy.$

Substituting $u = \log_b c$ gives

$x = y , \log_b c.$

By the definition of $x$, this proves that

$\log_b(c^y) = y , \log_b c.$

This completes the proof.

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