Introduction to Logarithms: Practice

Sums of logs become a product, and the subtracted log becomes a division.

$\log_{b}(x) + \log_{b}(y) - \log_{b}(z) = \log_{b}\left(\frac{xy}{z}\right)$

Change of base divides the log of the input by the log of the base.

$\log_{2}(7) = \frac{\log(7)}{\log(2)}$

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

$x\log(5) = \log(40) \;\Rightarrow\; x = \frac{\log(40)}{\log(5)}$

Since $10 = 2 \cdot 5$, the product law adds the two logs.

$\log_{b}(10) = \log_{b}(2) + \log_{b}(5) = 0.43 + 1 = 1.43$

Since $8 = 2^{3}$, the power law multiplies the log by $3$.

$\log_{b}(8) = 3\log_{b}(2) = 3(0.43) = 1.29$

Take the log of both sides and solve for the exponent.

$t\log(1.10) = \log(2) \;\Rightarrow\; t = \frac{\log(2)}{\log(1.10)}$

The product law recombines the two logs, and the factors of $3$ cancel.

$\log_{2}(3) + \log_{2}\left(\tfrac{8}{3}\right) = \log_{2}\left(3 \cdot \tfrac{8}{3}\right) = \log_{2}(8) = 3$

The coefficient goes back up as an exponent by the power law.

$2\log_{b}(3) = \log_{b}(3^{2}) = \log_{b}(9)$

A coefficient of $\tfrac{1}{2}$ becomes an exponent of $\tfrac{1}{2}$, which is a square root.

$\tfrac{1}{2}\log_{b}(x) = \log_{b}\left(x^{1/2}\right) = \log_{b}(\sqrt{x})$

A function and its inverse are mirror images across the diagonal.

$y = \log_{b}(x) \text{ reflects } y = b^{x} \text{ across } y = x$

Apply the power law first, then the product law.

$3\log_{b}(2) + \log_{b}(5) = \log_{b}(8) + \log_{b}(5) = \log_{b}(8 \cdot 5) = \log_{b}(40)$

Since $10$ lies between $8 = 2^{3}$ and $16 = 2^{4}$, the log is between $3$ and $4$, and $10$ is nearer $8$.

$8 < 10 < 16 \;\Rightarrow\; 3 < \log_{2}(10) < 4 \approx 3.32$

The nearest whole number is $3$.