6.3 Logarithms and the laws of logarithms — Practise with help

1. $2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$.
2. $5^0 = 1$ and $9^{1/2} = 3$.
3. $\left(e^{x}\right)^2 = e^{2x}$.

If question 1 was hard, revise negative indices in Chapter 2 before you continue.

The base of the power is the base of the logarithm. The exponent becomes the value of the logarithm.

The base is $3$, the exponent is $4$, so write $\log_3 81 = 4$.

The logarithm counts the number of factors of $3$ in $81$.

Write $\dfrac{1}{16}$ as a power of $2$, using a negative index.

$\dfrac{1}{16} = 2^{-4}$, so the logarithm is the exponent.

A negative logarithm means the input is smaller than $1$.

Adding logarithms with the same base multiplies the inputs.

Combine $5$ and $7$ under one logarithm with base $3$.

The product law is the reverse of the index law $3^m \times 3^n = 3^{m+n}$.

First use the product law, then use the power law on each logarithm.

Split the product into $\ln x^2 + \ln y^3$, then bring each power to the front.

Each power becomes a multiplier of its own logarithm.

Rewrite the logarithm as a power with base $4$.

Write $x = 4^3$ and work out the value.

The answer is positive, so it lies in the domain.

A negative value on the right is the exponent, not the input. The input $x$ is still $2^{-2}$.

Write $x = 2^{-2}$ and simplify it as a fraction.

The value $\dfrac{1}{4}$ is positive, so it is allowed. The negative sign sits in the exponent and does not make the input negative.

Let $y = \log_5 12$, write it as $5^y = 12$, then take the natural logarithm of both sides.

Taking logs gives $y\ln 5 = \ln 12$. Make $y$ the subject.

The base of the original logarithm becomes the denominator.

Work out both sides with $x = y = 10$, using $\ln 20$ on the left and $\ln 10 + \ln 10$ on the right.

Use a calculator to find $\ln 20$ and $2\ln 10$, and compare them.

The two sides are different. The product law applies to a product, not a sum: $\ln(xy) = \ln x + \ln y$, but there is no such rule for $\ln(x + y)$.