# 4.6.2 Indices and Logarithms, SPM Practice (Long Questions)

4.6.2 Indices and Logarithms, SPM Practice (Long Questions)

Question 3:
Given that = 3r and q = 3t, express the following in terms of and/ or t.
(a)
(b)  log9p – log27 q.

Solution:
(a)
Given p = 3r, log3 p = r
q= 3t, log3 q =t

${\mathrm{log}}_{3}\frac{p{q}^{2}}{27}$
= log3 pq2 – log327
= log3 p + log3 q2 – log3 33
= r + 2 log3 q – 3 log3 3
= r + 2 log3 q – 3(1)
= r + 2t – 3

(b)
log9 p– log27 q
$\begin{array}{l}=\frac{{\mathrm{log}}_{3}p}{{\mathrm{log}}_{3}9}-\frac{{\mathrm{log}}_{3}q}{{\mathrm{log}}_{3}27}\\ =\frac{r}{{\mathrm{log}}_{3}{3}^{2}}-\frac{t}{{\mathrm{log}}_{3}{3}^{3}}\\ =\frac{r}{2{\mathrm{log}}_{3}3}-\frac{t}{3{\mathrm{log}}_{3}3}\\ =\frac{r}{2}-\frac{t}{3}\end{array}$

Question 4:
(a)  Simplify:
log2(2x + 1) – 5 log4 x2 + 4 log2 x
(b)  Hence, solve the equation:
log2(2x + 1) – 5 log4 x2 + 4 log2 x = 4

Solution:
(a)
log2 (2x + 1) – 5 log4 x2 + 4 log2 x
$\begin{array}{l}={\mathrm{log}}_{2}\left(2x+1\right)-\frac{5{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}4}+4{\mathrm{log}}_{2}x\\ ={\mathrm{log}}_{2}\left(2x+1\right)-\frac{5}{2}{\mathrm{log}}_{2}{x}^{2}+{\mathrm{log}}_{2}{x}^{4}\\ ={\mathrm{log}}_{2}\left(2x+1\right)-{\mathrm{log}}_{2}{\left({x}^{2}\right)}^{\left(\frac{5}{2}\right)}+{\mathrm{log}}_{2}{x}^{4}\end{array}$
$\begin{array}{l}={\mathrm{log}}_{2}\left(2x+1\right)-{\mathrm{log}}_{2}{x}^{5}+{\mathrm{log}}_{2}{x}^{4}\\ ={\mathrm{log}}_{2}\frac{\left(2x+1\right)\left({x}^{4}\right)}{{x}^{5}}\\ ={\mathrm{log}}_{2}\frac{2x+1}{x}\end{array}$

(b)
log2 (2x + 1) – 5 log4 x2 + 4 log2 x = 4