Properties of Logarithms: Practice

Split the fraction first with the quotient rule, which puts a minus sign in front of the denominator only.

$\log_b\!\left(\frac{x^2 y}{z}\right) = \log_b\!\left(x^2 y\right) - \log_b z$

The numerator is a product, so the product rule splits it, and the power rule then brings the exponent $2$ to the front.

$= 2\log_b x + \log_b y - \log_b z$

Only $z$ sits in the denominator, so only its logarithm is subtracted.

Both $27$ and $81$ are powers of $3$, so change to base $3$, where each logarithm is exact.

$\log_{27} 81 = \frac{\log_3 81}{\log_3 27} = \frac{4}{3}$

Check against the definition: $27^{4/3} = \left(3^3\right)^{4/3} = 3^4 = 81$. Correct.

The answer has to be a little more than $1$, because $81$ is only slightly larger than the base $27$. Dividing the arguments instead of their logarithms would give $3$, which is far too big.

For $x > 0$ the power rule applies directly, so both sides exist and are equal.

For $x < 0$ the left side is defined, since $x^2 > 0$, but the right side contains $\log_b x$ with a negative argument, which does not exist. At $x = 0$ neither side exists.

$\text{At } x = -4\colon \quad \log_b\!\left((-4)^2\right) = \log_b 16, \quad \text{while } 2\log_b(-4) \text{ is undefined.}$

So the statement holds exactly on $x > 0$. The version that survives for every $x \ne 0$ is $\log_b\!\left(x^2\right) = 2\log_b |x|$.

Rewrite the second factor in base $2$ using change of base, so that both logarithms live in the same base.

$\log_3 8 = \frac{\log_2 8}{\log_2 3}$

Now the product telescopes, because $\log_2 3$ cancels.

$\log_2 3 \cdot \frac{\log_2 8}{\log_2 3} = \log_2 8 = 3$

The inner base and the outer argument chain together: the $3$s cancel and only $\log_2 8$ survives.

Neither logarithm is a whole number alone, so combine them with the product rule and let the awkward numbers cancel.

$\log_3 18 + \log_3 \frac{3}{2} = \log_3\!\left(18 \cdot \frac{3}{2}\right) = \log_3 27 = 3$

Since $3^3 = 27$, the value is exactly $3$. Adding the arguments instead of multiplying would give $\log_3 \tfrac{39}{2}$, and dividing by $\tfrac{3}{2}$ instead of multiplying would give $\log_3 12$; neither is a whole number, which is a hint that the wrong operation was used.

Work inside the parentheses first. The power rule turns $2\log_b y$ into $\log_b\!\left(y^2\right)$, and the product rule then combines the sum.

$\log_b x + 2\log_b y = \log_b\!\left(xy^2\right)$

Now the outer coefficient $\tfrac{1}{3}$ goes up as an exponent, and the exponent $\tfrac{1}{3}$ is a cube root.

$\frac{1}{3}\log_b\!\left(xy^2\right) = \log_b\!\left(\left(xy^2\right)^{1/3}\right) = \log_b\!\left(\sqrt[3]{xy^2}\right)$

Three of the four are genuine identities. The power rule gives $2\log_b 3 = \log_b\!\left(3^2\right) = \log_b 9$; the quotient rule gives $\log_b 27 - \log_b 3 = \log_b\!\left(\tfrac{27}{3}\right) = \log_b 9$; and adding $\log_b 3$ to itself is the same as doubling it.

The quotient of two logarithms is a change of base, not a subtraction.

$\frac{\log_b 27}{\log_b 3} = \log_3 27 = 3$

That is the constant $3$, while $\log_b 9$ changes with $b$. With $b = 2$, for instance, $\log_2 9 \approx 3.17 \ne 3$. The two do happen to agree at the single base $b = \sqrt[3]{9}$, which is exactly why the word "always" belongs in the question: an identity has to hold for every admissible base, not just for a lucky one.

The product rule splits the two factors, and a square root is the exponent $\tfrac{1}{2}$.

$\log_5\!\left(25\sqrt{x}\right) = \log_5 25 + \log_5\!\left(x^{1/2}\right)$

Since $5^2 = 25$, the first term is $2$, and the power rule brings $\tfrac{1}{2}$ to the front of the second.

$= 2 + \frac{1}{2}\log_5 x$

The root applies only to $x$, so the $\tfrac{1}{2}$ must not be distributed over the $2$.

Split the quotient, then bring the exponent down.

$\log_b\!\left(\frac{x^2}{b}\right) = \log_b\!\left(x^2\right) - \log_b b = 2\log_b x - \log_b b$

Every base satisfies $\log_b b = 1$, so substitute both known values.

$= 2(6) - 1 = 11$

Change of base sends the argument to the numerator and the old base to the denominator.

$\log_7 x = \frac{\log_{10} x}{\log_{10} 7}$

Test it on a value you can check. With $x = 49$, the formula gives $\dfrac{\log_{10} 49}{\log_{10} 7} = \dfrac{2\log_{10} 7}{\log_{10} 7} = 2$, which is right because $7^2 = 49$. The upside-down version would give $\tfrac{1}{2}$.

The product rule extends to any number of terms, since you can apply it twice.

$\log_6 2 + \log_6 3 + \log_6 6 = \log_6(2 \cdot 3 \cdot 6) = \log_6 36$

Since $6^2 = 36$, the value is $2$. Adding the arguments instead would give $\log_6 11$, which is not even a whole number.

A sum of logarithms is the logarithm of the product, so multiply the two arguments.

$\log_b(x + 2) + \log_b(x - 2) = \log_b\big((x + 2)(x - 2)\big) = \log_b\!\left(x^2 - 4\right)$

The product is a difference of squares. Both factors are positive because $x > 2$, which is exactly the condition the product rule needs.