Quick questions on Laws of logarithms explained: O-Level A-Maths

The logarithm is the inverse of an index:

Two values fall straight out of the definition:

To evaluate a logarithm in a base your calculator does not have, change the base to one it does (such as $10$ or $e$):

Most questions need the laws together: bring powers down first, then merge products into sums and quotients into differences, working towards a single logarithm or a numerical value. A common target form is $\log_a$ of a single simplified number, from which a value follows at once.

A frequent A-Maths task gives you $\log_a 2 = p$ and $\log_a 3 = q$ and asks for the logarithm of some related number. The method is to factorise that number into powers of $2$ and $3$, then apply the laws to break the logarithm into the given pieces. For $\log_a 12$, write $12 = 2^2 \times 3$, so $\log_a 12 = 2\log_a 2 + \log_a 3 = 2p + q$. Even fractions work: $\log_a 1.5 = \log_a \tfrac{3}{2} = q - p$.

Because a logarithm only accepts a positive argument, every solution to a logarithmic equation must be checked against the domain, and invalid roots discarded. After combining $\log_2 x + \log_2(x - 2) = 3$ into a quadratic with roots $x = 4$ and $x = -2$, only $x = 4$ survives, because $x = -2$ would make both $\log_2 x$ and $\log_2(x - 2)$ undefined. The reliable habit is to state the required domain ($x > 2$ here) before solving, so any candidate outside it is rejected on sight rather than overlooked. This domain check is where method marks are commonly lost.

Simplify $\log_5 50 - \log_5 2$. [2 marks]

Express $\log_2 24$ in terms of $\log_2 3$. [2 marks]

Use change of base to evaluate $\log_4 64$. [2 marks]