Dedicated to the Power Rule, which allows an exponent in the argument of a logarithm to be moved to the front as a coefficient. Covers application in both directions.
For any valid base \(b\) and any real number \(p\), \(\log_b(M^p) = p \cdot \log_b(M)\). The exponent on the argument can be brought out front as a multiplier.
Use the Power Rule to expand the expression \(\log_5(x^7)\).
$$7\log_5(x)$$
Use the Power Rule to expand \(\ln(\sqrt{y})\).
First, rewrite the square root as an exponent: \(\ln(y^{\frac{1}{2}})\). Then apply the power rule: \(\frac{1}{2}\ln(y)\).
Use the Power Rule to expand \(\log(\frac{1}{z^3})\).
First, rewrite the expression with a negative exponent: \(\log(z^{-3})\). Then apply the power rule: \(-3\log(z)\).
Condense the expression \(4\log_b(x)\) into a single logarithm.
$$\log_b(x^4)$$
Condense the expression \(-2\log_3(a)\) into a single logarithm.
$$\log_3(a^{-2}) \text{ or } \log_3(\frac{1}{a^2})$$
Condense the expression \(\frac{1}{3}\ln(8)\) into a single logarithm and simplify.
The expression becomes \(\ln(8^{\frac{1}{3}})\). Since the cube root of \(8\) is \(2\), the simplified answer is \(\ln(2)\).
Which property would you apply first to expand \(\log( (3x)^2 )\)?
The Power Rule. You would rewrite it as \(2 \log(3x)\). Afterwards, you would apply the Product Rule to get \(2(\log(3) + \log(x))\).
