The change of base rule for logarithms states that \(\log_b a = \dfrac{\log_c a}{\log_c b}\). Show that, for \(x \gt 0\), \(\log_9 x = \dfrac{1}{2}\log_3 x\...

Parts (a) and (b) work together: two logarithms in different bases cannot be combined, so the change of base rule is used to bring them both to base \(3\) before anything else happens. Part (c) then returns to a single base and tests the domain of a logarithm instead.

(a) Showing that \(\log_{9}x = \dfrac{1}{2}\log_{3}x\) [2 marks]

Apply the quoted rule \(\log_{b}a = \dfrac{\log_{c}a}{\log_{c}b}\) with \(b = 9\) and the new base \(c = 3\), chosen because \(9\) is a power of \(3\) and the denominator will therefore be a whole number:

\[\log_{9}x = \frac{\log_{3}x}{\log_{3}9} \qquad \textbf{M1}\]

The M1 is for using the rule with base \(3\), the useful choice. Since \(9 = 3^{2}\), we have \(\log_{3}9 = 2\), so

\[\log_{9}x = \frac{\log_{3}x}{2} = \frac{1}{2}\log_{3}x \qquad \textbf{A1}\]

as required. The result is printed, so the marks are for the two steps. The commonest error is inverting the rule and writing \(\dfrac{\log_{3}9}{\log_{3}x}\); a quick numerical test kills that version, since with \(x = 9\) the left-hand side is \(\log_{9}9 = 1\) while the correct right-hand side is \(\tfrac{1}{2}\log_{3}9 = \tfrac{1}{2} \times 2 = 1\). The second error is taking \(\log_{3}9\) to be \(3\) or \(9\); it is the power to which \(3\) must be raised to give \(9\), which is \(2\).

(b) Solving \(\log_{3}x + \log_{9}x = 6\) [3 marks]

The two terms are in different bases, so they cannot be added as they stand. Part (a) converts the second one:

\[\log_{3}x + \tfrac{1}{2}\log_{3}x = 6 \qquad \textbf{M1}\]

Now both terms are multiples of the same quantity, so they collect like ordinary algebra, treating \(\log_{3}x\) as a single unknown:

\[\tfrac{3}{2}\log_{3}x = 6 \quad \Longrightarrow \quad \log_{3}x = 6 \times \tfrac{2}{3} = 4 \qquad \textbf{M1}\] \[x = 3^{4}, \quad \text{that is } x = 81 \qquad \textbf{A1}\]

The question asks for the answer as a power of \(3\), so \(3^{4}\) is the requested form and \(81\) is its value. The wrong turns here are adding the two original logarithms as though the bases matched, giving \(2\log_{3}x = 6\) and the answer \(3^{3} = 27\), and multiplying \(6\) by \(\tfrac{3}{2}\) instead of dividing, which gives \(\log_{3}x = 9\). Check the answer in the original equation: \(\log_{3}81 = 4\) and \(\log_{9}81 = 2\), and \(4 + 2 = 6\).

(c) Solving \(\log_{3}y + \log_{3}\left(y - 8\right) = 2\) [3 marks]

Both logarithms already share the base \(3\), so the addition law combines them into the logarithm of a product, after which the logarithm can be removed:

\[\log_{3}\left[y\left(y - 8\right)\right] = 2 \quad \Longrightarrow \quad y\left(y - 8\right) = 3^{2} = 9 \qquad \textbf{M1}\]

Expand and factorise, seeking two numbers with product \(-9\) and sum \(-8\):

\[y^{2} - 8y - 9 = 0 \quad \Longrightarrow \quad \left(y - 9\right)\left(y + 1\right) = 0 \quad \Longrightarrow \quad y = 9 \text{ or } y = -1 \qquad \textbf{M1}\]

Now the justification the question demands. A logarithm is defined only for a strictly positive argument, so \(\log_{3}y\) requires \(y \gt 0\) and \(\log_{3}\left(y - 8\right)\) requires \(y \gt 8\); together the true domain is \(y \gt 8\). The value \(y = -1\) fails both, since neither \(\log_{3}\left(-1\right)\) nor \(\log_{3}\left(-9\right)\) exists, so it must be rejected:

\[y = 9 \qquad \textbf{A1}\]

The A1 needs the reason stated as well as the value, because the question says "justifying the rejection". Note why the extra root appeared: the product \(y\left(y - 8\right)\) is positive when both factors are negative as well as when both are positive, but the original logarithms require each factor to be positive on its own. That is why every root must be tested in the original equation and not merely in the quadratic. Checking the survivor: \(\log_{3}9 + \log_{3}1 = 2 + 0 = 2\), as required.

It is instructive to compare parts (b) and (c). In part (b) no root was rejected because \(3^{4}\) is automatically positive and there was only one logarithm's argument to consider. In part (c) a second argument, \(y - 8\), raises the floor of the domain from \(0\) to \(8\), and that is what condemns the negative root.

Examination takeaway. When two logarithms have different bases, convert to the base that makes the denominator a whole number, which is usually the smaller base. Then treat the logarithm as a single unknown and collect terms. Whenever logarithms are combined before being removed, work out the domain from every argument in the original equation and use it to justify rejecting any root that falls outside.