Properties of Logarithms: Additional Practice (Set 1)

Change the second factor to base $3$ so that both logarithms share a base.

$\log_5 9 = \frac{\log_3 9}{\log_3 5}$

The factor $\log_3 5$ now cancels, leaving a logarithm that is easy to evaluate.

$\log_3 5 \cdot \frac{\log_3 9}{\log_3 5} = \log_3 9 = 2$

The inner base $5$ has disappeared, which is the whole point of the chain: only the outer base and the final argument survive.

Rewrite $3.5$ with the numbers you have: $3.5 = \dfrac{7}{2}$.

The quotient rule turns the division into a subtraction.

$\log_b 3.5 = \log_b 7 - \log_b 2 \approx 0.98 - 0.35 = 0.63$

Adding gives $1.33$, which is $\log_b 14$, and multiplying the two logarithms gives $0.343$, which means nothing at all.

A quotient of two logarithms is not a difference of logarithms. It is a change of base.

$\frac{\log_b M}{\log_b N} = \log_N M$

Test it with $b = 10$, $M = 100$, $N = 10$: the left side is $\dfrac{2}{1} = 2$, while $\log_{10} 100 - \log_{10} 10 = 2 - 1 = 1$. The two disagree, so the first statement is false. The other three are the power, quotient, and product rules, stated with the domain conditions they need.

Both $16$ and $8$ are powers of $2$, so change to base $2$, where each logarithm is a whole number.

$\log_{16} 8 = \frac{\log_2 8}{\log_2 16} = \frac{3}{4}$

Check it against the definition: $16^{3/4} = \left(2^4\right)^{3/4} = 2^3 = 8$. Correct. The answer must be less than $1$, since $8$ is smaller than the base $16$.

Take base-10 logarithms and use the power rule to bring the exponent down.

$\log_{10}\!\left(3^{20}\right) = 20\log_{10} 3 \approx 20(0.4771) = 9.542$

That lies between $9$ and $10$, so the number lies between $10^9$ and $10^{10}$.

$10^9 < 3^{20} < 10^{10}$

An integer of at least $10^9$ but less than $10^{10}$ has exactly $10$ digits. (Indeed $3^{20} = 3{,}486{,}784{,}401$.)

A quotient of logarithms in a common base is a change of base, with the bottom argument becoming the new base.

$\frac{\log_b 81}{\log_b 3} = \log_3 81 = 4$

Since $3^4 = 81$, the value is $4$ for every valid $b$, so the original base cancels out completely. The power rule shows the same thing: $\log_b 81 = 4\log_b 3$, and the $\log_b 3$ divides away.

Write the decimal as a fraction built from the numbers you know: $0.75 = \dfrac{3}{4} = \dfrac{3}{2^2}$.

The quotient rule splits it, and the power rule handles the $2^2$.

$\log_b 0.75 = \log_b 3 - 2\log_b 2 = n - 2m$

A sanity check on the sign: $0.75 < 1$, so its logarithm is negative for any base $b > 1$, and indeed $n - 2m$ comes out negative there (for instance $b = 10$ gives $0.477 - 0.602 = -0.125$).

Combine all three with the product rule before evaluating anything, and let the $3$s cancel.

$\log_2 3 + \log_2 \frac{4}{3} + \log_2 2 = \log_2\!\left(3 \cdot \frac{4}{3} \cdot 2\right) = \log_2 8$

Since $2^3 = 8$, the value is $3$. None of the three logarithms is a whole number on its own except the last, which is the point: the rule lets the awkward parts cancel first.

Condense the exponent first with the product rule.

$\log_b 5 + \log_b 2 = \log_b(5 \cdot 2) = \log_b 10$

Now raising $b$ to a base-$b$ logarithm undoes it, because the two operations are inverses.

$b^{\,\log_b 10} = 10$

Adding the exponents corresponds to multiplying the numbers, which is the product rule read in the exponential direction.

Write the argument as a single power of $5$: $\dfrac{1}{125} = \dfrac{1}{5^3} = 5^{-3}$.

The power rule brings the negative exponent straight out in front.

$\log_b\!\left(5^{-3}\right) = -3\log_b 5 = -3(0.7) = -2.1$

The sign is a useful check: for $b > 1$ the logarithm of a number smaller than $1$ must be negative.

Change $\log_4 9$ to base $2$, since both $4$ and $9$ are perfect squares.

$\log_4 9 = \frac{\log_2 9}{\log_2 4} = \frac{2\log_2 3}{2} = \log_2 3$

The power rule turned $\log_2 9$ into $2\log_2 3$, and the $2$ in the denominator cancelled it. Numerically both sides are about $1.585$, while $\log_2 9 \approx 3.17$ and $\log_4 3 \approx 0.79$.

Change the second factor to base $b$, using $\log_b b = 1$.

$\log_x b = \frac{\log_b b}{\log_b x} = \frac{1}{\log_b x}$

The two factors are therefore reciprocals of each other, and their product is $1$.

$\left(\log_b x\right) \cdot \frac{1}{\log_b x} = 1$

The division is legal because $\log_b x \ne 0$, which is exactly what the condition $x \ne 1$ guarantees. Concretely, $\log_2 8 = 3$ and $\log_8 2 = \tfrac{1}{3}$, and their product is $1$.

A product of logarithms is never the logarithm of a product, which rules out $\log_b(bx)$, and it is not a sum either. The answer is a plain number, free of both $b$ and $x$.