Power Rule

The power rule of logarithms states that

\[ log(a^b) = b \cdot log(a) \]

Proof

Given that

\[ log_b(a) = x \]

It follows that

\[ b^x = a \]

Next, introduce another exponent \( c \) so both sides of this equation.

\[ (b^x)^c = a^c \]

\[ b^{x \cdot c} = a^c \]

This is essentially saying that if we raise \( b \) to \( x \cdot c \) we get \( a^c \). Another way of saying exactly that is using a \( log \) function with base \( b \). This is what log functions tell you by definition! So,

\[ log_b(a^c) = x \cdot c \]

We can now substitute into this equation, replacing \( x \) with \( log_b(a) \) as we stated above that \( log_b(a) = x \).

\[ log_b(a^c) = log_b(a) \cdot c \]

Re-arranged to

\[ log_b(a^c) = c \cdot log_b(a) \]

Which proves the power rule.

This makes sense intuitively because whenever you take a log of a single value you are raising some base to that value. If you raise that value itself to an exponent that just involves raising the base again. And, as we can see from above, doing an exponent of an exponent results in multiplication.

Supplementary Proof

This was the previous proof that I had for this that was essentially summarised from a Khan explanation. I’ve left it in for completeness.

Given that

\[ log_x(a) = b \]

It follows that

\[ x^b = a \]

Next we introduce a varaible \( c \) and apply it to the first equation above by multiplying both sides

\[ (*) \ c \cdot log_x(a) = c \cdot b \]

Then we can introduce \( c \) to the second equation by raising both sides to \( c \)

\[ (x^b)^c = a^c \]

\[ x^{b \cdot c} = a^c \]

We can rewrite this expression in terms of a logarithm of \( x \) - what do we raise \( x \) to to equal \( a^c \)? Well, its stated write there, as \( b \cdot c \).

\[ log_x(a^c) = b \cdot c \]

Above we can see that the \( (*) \) star figure shows

\[ c \cdot log_x(a) = c \cdot b \]

Therefore we just proved that

\[ c \cdot log_x(a) = log_x(a^c) \]

Because they both equal \( c \cdot b \).