The Power Rule for Logarithms

We have already seen that the logarithm of a product is the sum of the logarithms of the factors:

 $\displaystyle \log _b \left( {xy} \right) = \log _b \left( x \right) + \log _b \left( y \right)$

If we apply this rule repeatedly we can devise another rule for simplifying expressions of the form $\log_b x^p$. 

Recall that $x^p$ can be thought of as $x \cdot x \cdot x \cdots x$ where there are $p$ factors of $x$. Then we have:

$\displaystyle \begin{aligned} \log_b(x^p) &= \log_b (x \cdot x \cdots x) \\ &= \log_b x + \log_b x + \cdots +\log_b x \\ &= p\log_b x \end{aligned}$ 

Since the $p$ factors of $x$ are converted to $p $ summands by the product rule formula.  

Example 1: Simplify the expression $\log_3(3^x\cdot 9x^{100})$

First expand the log:

$\displaystyle \log_3(3^x\cdot 9x^{100}) =\log_3 (3^x) + \log_3 9 + \log_3(x^{100}) $ 

Next use the product and power rule to simplify:

$\displaystyle \log_3 (3^x) + \log_3 9 + \log_3 (x^{100})= x+2+100\log_3 x$

Example 2: Solve $2^{(x+1)}=10^3$ for $x$ using logarithms

Start by taking the logarithm with base $2$ of both sides:

$\displaystyle \begin{aligned} \log_2 (2^{(x+1)}) &= \log_2 (10^3)\\ x+1&=3\log_2(10)\\ x&=3\log_2(10)-1 \end{aligned}$ 

Therefore a solution would be $x=3\log_2(10) -1. $