The Logarithm Power Rule

The Logarithm Power Rule comes in handy when we have an exponent inside the logarithm.

The rule is:

\[\log_b(A^B) = B \cdot \log_b(A)\]

---

1. The Simple Case (Intuition)

Let’s first see the pattern with a simple example where the log’s base (\(b\)) is the same as the argument’s base (\(A\)).

Let \(b = 10\), \(A = 10\), and \(B = 2\):

\[\log_{10}(10^2)\]

This is just \(\log_{10}(100)\). The question is, “What power do I raise 10 to, to get 100?” The answer is 2.

So, \(\log_{10}(10^2) = 2\). The answer is simply the exponent \(B\). The base and the log “cancel out.”

2. The General Case (Proof)

But what happens when the log’s base (\(b\)) is different from the argument’s base (\(A\))?

Let \(b = 10\), \(A = 5\), and \(B = 2\):

\[\log_{10}(5^2)\]

This is not as straightforward. We can’t just “cancel” the base. This is where the full rule is needed. Let’s prove it.

---

The Proof

We want to prove that \(\log_b(A^B) = B \cdot \log_b(A)\).

Step 1: We know that \(\log_b(A)\) evaluates to a single number, which is the exponent. Let’s call this resulting exponent \(M\)

\[M = \log_b(A)\]

Step 2: By the fundamental definition of a logarithm, this means:

\[b^M = A\]

Step 3: Now, let’s look at the left side of our main equation, \(\log_b(A^B)\), and substitute \(A\) with \(b^M\):

\[\log_b( (b^M)^B )\]

Step 4: Using the exponent rule \((x^m)^n = x^{m \cdot n}\), we simplify the inside:

\[\log_b( b^{M \cdot B} )\]

Step 5: Now we are back to the “simple case.” The expression asks, “What power do I raise \(b\) to, to get \(b^{M \cdot B}\)?” The answer is just the exponent:

\[M \cdot B\]

Step 6: We have now proven that the left side, \(\log_b(A^B)\), is equal to \(M \cdot B\). Now, let’s substitute \(M\) (from Step 1) back into our result:

\[B \cdot \log_b(A)\]

Conclusion: We have shown that \(\log_b(A^B)\) simplifies to \(B \cdot \log_b(A)\).

Hence:

\[\log_{b}(A^B) = B \log_{b}(A)\]