(a) (i) Given that \(\log_8 x^3 = \log_4 u\), we need to express \(u\) in terms of \(x\).
First, convert the logarithms to base 2:
\( \log_8 x^3 = \frac{\log_2 x^3}{\log_2 8}\)
\( = \frac{3 \log_2 x}{3} = \log_2 x \)
\( \log_4 u = \frac{\log_2 u}{\log_2 4} = \frac{\log_2 u}{2} \)
Setting these equal:
\( \log_2 x = \frac{\log_2 u}{2} \)
Multiplying both sides by 2:
\( 2 \log_2 x = \log_2 u \)
\(\log_2 x^2 = \log_2 u \)
\( u = x^2 \)
Thus, the answer is \(u = x^2\).
(ii) We need to solve the equation \(\log_4(x^2 + 5x) – \log_8 x^3 = \frac{1}{\log_3 4}\).
Using the answer from part (i)
\(\log_8 x^3 = \log_4 x^2\)
Also \( \frac{1}{\log_3 4} = \log_4 3 \)
\(\log_4(x^2 + 5x) – \log_4 x^2 = \log_4 3 \)
\( \log_4\left(\frac{x^2 + 5x}{x^2}\right) = \log_4 3 \)
\( \frac{x^2 + 5x}{x^2} = 3 \)
Solve for \(x\):
\( 1 + \frac{5}{x} = 3 \)
\( \frac{5}{x} = 2 \)
\( x = \frac{5}{2} \)
Thus, the answer is \(\frac{5}{2}\).
(b) Solve the equation \(e^y(e^y – 2) = 15\).
Let \(z = e^y\), then the equation becomes:
\( z(z – 2) = 15 \)
\( z^2 – 2z – 15 = 0 \)
Solving the quadratic equation:
\( z = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2} \)
\(z = 5 \text{ or } z = -3 \)
Since \(z = e^y\) must be positive, we have \(z = 5\).
Thus: \( e^y = 5 \)
\(y = \ln 5 \)
Thus, the answer is \(\ln 5\).
