Introduction to Logarithms: Additional Practice (Set 2)

The quotient law splits the fraction, then the power law drops the exponent $3$ in front.

$\log_{b}\left(\frac{x^{3}}{y}\right) = \log_{b}(x^{3}) - \log_{b}(y) = 3\log_{b}(x) - \log_{b}(y)$

Rewrite each factor with common logs, then cancel the matching $\log(3)$.

$\frac{\log 3}{\log 2} \cdot \frac{\log 8}{\log 3} = \frac{\log 8}{\log 2} = \log_{2}(8) = 3$

Rewrite the logarithmic equation in exponential form.

$\log_{2}(x) = 5 \iff x = 2^{5} = 32$

Dividing the natural log of the input by the natural log of the base rewrites any logarithm in base $e$.

$\log_{b}(x) = \frac{\ln(x)}{\ln(b)}$

This is the change-of-base relationship.

Ask: eight to what power is $2$? A cube root is the power $\tfrac{1}{3}$.

$8^{1/3} = \sqrt[3]{8} = 2 \;\Rightarrow\; \log_{8}(2) = \tfrac{1}{3}$

Combine with the product law, rewrite in exponential form, and solve the quadratic.

$\log_{2}\big(x(x-2)\big) = 3 \;\Rightarrow\; x^{2} - 2x = 8 \;\Rightarrow\; (x - 4)(x + 2) = 0$

Only $x = 4$ keeps both inputs positive, so $x = 4$.

Write $\tfrac{1}{16}$ as a power of $\tfrac{1}{2}$ and match the exponents.

$\left(\tfrac{1}{2}\right)^{t} = \frac{1}{16} = \left(\tfrac{1}{2}\right)^{4} \;\Rightarrow\; t = 4$

The number of digits of a whole number $N$ is $\lfloor \log N\rfloor + 1$.

$\log(2^{100}) = 100\log(2) \approx 30.1 \;\Rightarrow\; \lfloor 30.1\rfloor + 1 = 31$

Rewrite both numbers in base $2$: $4^{x} = 2^{2x}$ and $32 = 2^{5}$.

$2x = 5 \;\Rightarrow\; x = \tfrac{5}{2}$

Rewrite in exponential form, then solve the linear equation.

$2x + 1 = 3^{2} = 9 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4$

The logarithm inverts the exponential, whose domain is all real numbers, so the log's range is all real numbers.

$\text{range: all real numbers}$

The common log has base $10$, so rewrite in exponential form.

$\log(x) = 2 \iff x = 10^{2} = 100$