Properties of Logarithms: Practice

The product rule says that a product inside a logarithm becomes a sum outside it, because a logarithm is an exponent and exponents add when powers multiply.

$\log_b(7 \cdot 5) = \log_b 7 + \log_b 5$

Multiplying or dividing the two logarithms is not a rule at all. A quotient of logarithms, $\dfrac{\log_b 7}{\log_b 5}$, is the change-of-base formula and equals $\log_5 7$.

A difference of logarithms with the same base is the logarithm of the quotient, so divide the arguments rather than subtracting them.

$\log_5 12 - \log_5 4 = \log_5\!\left(\frac{12}{4}\right) = \log_5 3$

Subtracting the arguments would give $\log_5 8$, which is a different number: the quotient rule divides, it does not subtract.

Each logarithm is a whole number here, so evaluate them directly: $2^3 = 8$ gives $\log_2 8 = 3$, and $2^2 = 4$ gives $\log_2 4 = 2$.

$\log_2 8 + \log_2 4 = 3 + 2 = 5$

The product rule gives the same answer by another route: $\log_2 8 + \log_2 4 = \log_2 32 = 5$, since $2^5 = 32$.

Ask what exponent turns $b$ into $1$. Any nonzero base raised to the power $0$ gives $1$.

$b^0 = 1 \quad\Longrightarrow\quad \log_b 1 = 0$

This is the fact behind the reciprocal shortcut $\log_b\!\left(\tfrac{1}{N}\right) = \log_b 1 - \log_b N = -\log_b N$.

Ask what exponent turns $7$ into $7$. The first power of a number is the number itself.

$7^1 = 7 \quad\Longrightarrow\quad \log_7 7 = 1$

In general $\log_b b = 1$ for every valid base.

Neither logarithm is a whole number by itself, so combine them first with the product rule.

$\log_6 4 + \log_6 9 = \log_6(4 \cdot 9) = \log_6 36$

Now ask what exponent turns $6$ into $36$. Since $6^2 = 36$, the value is $2$. Adding the arguments instead of multiplying them would give $\log_6 13$, which is wrong.

Change of base puts the argument on top and the old base underneath.

$\log_5 20 = \frac{\log_{10} 20}{\log_{10} 5} \approx \frac{1.301}{0.699} \approx 1.861$

Check the size: $5^1 = 5$ and $5^2 = 25$, and $20$ lies between them, so the answer must lie between $1$ and $2$. The upside-down version would give about $0.537$.

Only the product rule is a genuine identity: a product inside becomes a sum outside.

$\log_b(MN) = \log_b M + \log_b N$

The two statements about $\log_b(M + N)$ die on a single example. Take $b = 2$ and $M = N = 4$, so that $\log_2(4 + 4) = \log_2 8 = 3$, while $\log_2 4 + \log_2 4 = 4$ and $\log_2 4 \cdot \log_2 4 = 4$. A logarithm never splits across a sum.

Multiplying logarithms is not a rule either. Take $b = 2$ and $M = N = 2$.

$\log_2(2 \cdot 2) = 2, \quad \text{but} \quad \log_2 2 \cdot \log_2 2 = 1 \cdot 1 = 1$

The power rule pulls the exponent to the front, and $\log_5 5 = 1$.

$\log_5\!\left(5^7\right) = 7\log_5 5 = 7(1) = 7$

This is the inverse relationship in action: taking a base-5 logarithm undoes raising $5$ to a power, so $\log_b\!\left(b^x\right) = x$ for every real $x$.

Combine the two logarithms with the product rule before evaluating anything.

$\log_4 2 + \log_4 32 = \log_4(2 \cdot 32) = \log_4 64$

Now ask what exponent turns $4$ into $64$. Since $4^3 = 64$, the value is $3$.

Send the coefficient up as an exponent with the power rule, remembering that the exponent $\tfrac{1}{2}$ means a square root.

$\frac{1}{2}\log_b 25 = \log_b\!\left(25^{1/2}\right) = \log_b \sqrt{25} = \log_b 5$

The coefficient acts on the argument as an exponent, not as a divisor, so halving the argument to get $\log_b 12.5$ is wrong.

The product rule splits the two factors, then the power rule brings the exponent down in front of the first one.

$\log_b\!\left(x^3 y\right) = \log_b\!\left(x^3\right) + \log_b y = 3\log_b x + \log_b y$

The exponent $3$ belongs only to $x$, so it multiplies only $\log_b x$. Distributing it across both terms, as in $3\left(\log_b x + \log_b y\right)$, would be the logarithm of $x^3 y^3$.