In this question all logarithms are to the base stated, and \(a\), \(b\) and \(x\) are real numbers. Given that \(\log_{2}a + \log_{2}b = 5\) and \(\log_{2}...

The three parts exercise three different uses of logarithms: treating logarithms as unknowns in a pair of simultaneous equations, combining logarithms into one before undoing them, and using a substitution to turn an exponential equation into a quadratic. All of them rest on the definition \(\log_{2} N = m\) meaning \(N = 2^{m}\).

(a) Finding \(a\) and \(b\) [4 marks]

Do not expand the logarithms into a product. Treat \(\log_{2}a\) and \(\log_{2}b\) as two unknowns in a linear system, which is what makes elimination the right method. Adding the two equations removes \(\log_{2}b\):

\[\left(\log_{2}a + \log_{2}b\right) + \left(\log_{2}a - \log_{2}b\right) = 5 + 1 \quad \Longrightarrow \quad 2\log_{2}a = 6 \quad \Longrightarrow \quad \log_{2}a = 3 \qquad \textbf{M1 A1}\]

The M1 is for a correct elimination and survives an arithmetic slip; the A1 needs the value \(3\). Converting from logarithmic to index form:

\[a = 2^{3} = 8 \qquad \textbf{A1}\]

Subtracting the equations instead removes \(\log_{2}a\):

\[2\log_{2}b = 5 - 1 = 4 \quad \Longrightarrow \quad \log_{2}b = 2 \quad \Longrightarrow \quad b = 2^{2} = 4 \qquad \textbf{A1}\]

Check in the original statements: \(\log_{2}8 + \log_{2}4 = 3 + 2 = 5\) and \(3 - 2 = 1\), both correct. The characteristic error is stopping at \(\log_{2}a = 3\) and offering that as the answer for \(a\); the question asks for \(a\) itself, so the conversion to \(8\) is required. A second error is dividing \(6\) by \(2\) after taking a logarithm of both sides, which confuses \(2\log_{2}a\) with \(\log_{2}\left(2a\right)\).

(b) Solving \(\log_{2}\left(3x + 2\right) - \log_{2}\left(x - 1\right) = 3\) [4 marks]

A difference of two logarithms with the same base is the logarithm of a quotient, so combine them into a single logarithm. That is the step that allows the logarithm to be removed:

\[\log_{2}\left(\frac{3x + 2}{x - 1}\right) = 3 \qquad \textbf{M1}\] \[\frac{3x + 2}{x - 1} = 2^{3} = 8 \qquad \textbf{M1}\]

The second M1 is for correctly undoing the logarithm; writing the right-hand side as \(3\), or as \(3^{2}\), is a common way to lose it. Multiply up and solve the linear equation:

\[3x + 2 = 8\left(x - 1\right) = 8x - 8 \quad \Longrightarrow \quad 10 = 5x \qquad \textbf{M1}\] \[x = 2 \qquad \textbf{A1}\]

A logarithm is defined only for a positive argument, so both \(3x + 2 \gt 0\) and \(x - 1 \gt 0\) are needed, which together require \(x \gt 1\). Since \(x = 2\) satisfies that, it is valid. Substituting back confirms it: \(\log_{2}8 - \log_{2}1 = 3 - 0 = 3\). The wrong turn here is subtracting the arguments rather than dividing them, that is writing \(\log_{2}\left(3x + 2 - x + 1\right)\), which destroys the equation.

(c) Solving \(2^{2x+1} - 17\left(2^{x}\right) + 8 = 0\) [5 marks]

The equation contains \(2^{x}\) and \(2^{2x+1}\), which look different but are related, so a substitution reduces it to a quadratic. Put \(y = 2^{x}\). The key manipulation is splitting the index using the laws of indices:

\[2^{2x+1} = 2^{1} \times 2^{2x} = 2 \times \left(2^{x}\right)^{2} = 2y^{2} \qquad \textbf{M1}\]

This is the mark the question is built around. Writing \(2^{2x+1}\) as \(y^{2}\), or as \(2y^{2x}\), or as \(\left(2^{x}\right)^{2x+1}\), all fail here. With that substitution:

\[2y^{2} - 17y + 8 = 0 \qquad \textbf{A1}\] \[\left(2y - 1\right)\left(y - 8\right) = 0 \quad \Longrightarrow \quad y = \frac{1}{2} \text{ or } y = 8 \qquad \textbf{M1 A1}\]

Now reverse the substitution. Because \(2^{x} \gt 0\) for every real \(x\), both roots are admissible, unlike the logarithm questions where a root is often rejected:

\[2^{x} = \frac{1}{2} = 2^{-1} \quad \Longrightarrow \quad x = -1; \qquad 2^{x} = 8 = 2^{3} \quad \Longrightarrow \quad x = 3 \qquad \textbf{A1}\]

Both must be given: offering only \(x = 3\) loses the final accuracy mark. Verify the awkward one directly. With \(x = -1\), \(2^{2x+1} = 2^{-1} = \tfrac{1}{2}\) and \(2^{x} = \tfrac{1}{2}\), so the expression is \(\tfrac{1}{2} - \tfrac{17}{2} + 8 = 0\). With \(x = 3\), \(2^{7} = 128\) and \(128 - 17\left(8\right) + 8 = 128 - 136 + 8 = 0\). The final trap is forgetting to return from \(y\) to \(x\) and quoting \(\tfrac{1}{2}\) and \(8\) as the answers.

Examination takeaway. Combine logarithms into a single one before removing them, and always convert a result such as \(\log_{2}a = 3\) into the value of \(a\). In an exponential equation, look for a term whose index is double another and split it with the index laws, remembering that a substituted exponential is always positive so neither root of the quadratic can be dismissed on those grounds.