Introduction to Logarithms: Additional Practice (Set 1)

Write $16$ as a power of $2$ and match the exponents.

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

Since $6 = 2 \cdot 3$, the product law adds the two logs.

$\log(6) = \log(2) + \log(3) \approx 0.301 + 0.477 = 0.778$

Divide by $3$ first to isolate the power, then match the exponents.

$2^{x} = 16 = 2^{4} \;\Rightarrow\; x = 4$

Turn each coefficient into an exponent, then combine additions as products and the subtraction as a quotient.

$\log(x) - 3\log(y) + \tfrac{1}{2}\log(z) = \log\left(\frac{x \cdot z^{1/2}}{y^{3}}\right) = \log\left(\frac{x\sqrt{z}}{y^{3}}\right)$

Take the log of both sides and bring the exponent down with the power law.

$t\log(3) = \log(90) \;\Rightarrow\; t = \frac{\log(90)}{\log(3)}$

Combine the left side with the product law, then match the inputs.

$\log_{5}(3x) = \log_{5}(12) \;\Rightarrow\; 3x = 12 \;\Rightarrow\; x = 4$

Rewrite both numbers in base $3$: $27^{x} = 3^{3x}$ and $9 = 3^{2}$.

$3x = 2 \;\Rightarrow\; x = \tfrac{2}{3}$

The equation says $b^{2} = 49$.

$b^{2} = 49 \;\Rightarrow\; b = 7$

The base must be positive, so $b = 7$.

Write $1000 = 10^{3}$ and match the exponents.

$10^{2x} = 10^{3} \;\Rightarrow\; 2x = 3 \;\Rightarrow\; x = \tfrac{3}{2}$

The quotient law subtracts the two logs.

$\log_{b}\left(\frac{x}{y}\right) = \log_{b}(x) - \log_{b}(y) = 5 - 2 = 3$

The power law handles $x^{2}$, then the product law adds.

$\log_{b}(x^{2}y) = 2\log_{b}(x) + \log_{b}(y) = 2(5) + 2 = 12$

Take the natural log of both sides; $\ln$ and $e$ undo each other.

$\ln(e^{x}) = \ln(20) \;\Rightarrow\; x = \ln(20)$

The common log $\log(20)$ is a different value, since it uses base $10$.