 # Indices and Logarithms, Short Questions (Question 1 – 8)

Question 1
Solve the equation, log3 [log2(2x – 1)] = 2

Solution:
log3 [log2 (2x – 1)] = 2 ← (if log a N = x, N = ax)
log2 (2x – 1) = 32
log2 (2x – 1) = 9
2x – 1 = 29
x = 256.5

Question 2
Solve the equation,

Solution:

Question 3
Solve the equation, ${5}^{{\mathrm{log}}_{4}x}=125$

Solution:

Question 4
Solve the equation, ${5}^{{\mathrm{log}}_{5}\left(x+1\right)}=9$

Solution:
$\begin{array}{l}{5}^{{\mathrm{log}}_{5}\left(x+1\right)}=9\\ {\mathrm{log}}_{5}{5}^{{}^{{\mathrm{log}}_{5}\left(x+1\right)}}={\mathrm{log}}_{5}9\\ {\mathrm{log}}_{5}\left(x+1\right).{\mathrm{log}}_{5}5={\mathrm{log}}_{5}9\\ {\mathrm{log}}_{5}\left(x+1\right)={\mathrm{log}}_{5}9\\ x+1=9\\ x=8\end{array}$

Question 5
Solve the equation, ${\mathrm{log}}_{9}\left(x-2\right)={\mathrm{log}}_{3}2$

Solution:
$\begin{array}{l}{\mathrm{log}}_{9}\left(x-2\right)={\mathrm{log}}_{3}2\\ \overline{){\mathrm{log}}_{a}b=\frac{{\mathrm{log}}_{c}b}{{\mathrm{log}}_{c}a}}⇒\frac{{\mathrm{log}}_{3}\left(x-2\right)}{{\mathrm{log}}_{3}9}={\mathrm{log}}_{3}2\\ \frac{{\mathrm{log}}_{3}\left(x-2\right)}{2}={\mathrm{log}}_{3}2\\ {\mathrm{log}}_{3}\left(x-2\right)=2{\mathrm{log}}_{3}2\\ {\mathrm{log}}_{3}\left(x-2\right)={\mathrm{log}}_{3}{2}^{2}\\ x-2=4\\ x=6\end{array}$

Question 6
Solve the equation, ${\mathrm{log}}_{9}\left(2x+12\right)={\mathrm{log}}_{3}\left(x+2\right)$

Solution:

Question 7
Solve the equation, ${\mathrm{log}}_{4}x=\frac{3}{2}{\mathrm{log}}_{2}3$

Solution:
$\begin{array}{l}{\mathrm{log}}_{4}x=\frac{3}{2}{\mathrm{log}}_{2}3\\ \frac{{\mathrm{log}}_{2}x}{{\mathrm{log}}_{2}4}=\frac{3}{2}{\mathrm{log}}_{2}3\\ \frac{{\mathrm{log}}_{2}x}{2}=\frac{3}{2}{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x=2×\frac{3}{2}{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x=3{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x={\mathrm{log}}_{2}{3}^{3}\\ x=27\end{array}$

Question 8
Solve the equation, $\frac{2}{{\mathrm{log}}_{5}2}={\mathrm{log}}_{2}\left(2-x\right)$

Solution:
$\begin{array}{l}\frac{2}{{\mathrm{log}}_{5}2}={\mathrm{log}}_{2}\left(2-x\right)\\ 2={\mathrm{log}}_{5}2.{\mathrm{log}}_{2}\left(2-x\right)\\ 2=\frac{1}{{\mathrm{log}}_{2}5}.{\mathrm{log}}_{2}\left(2-x\right)\\ 2{\mathrm{log}}_{2}5={\mathrm{log}}_{2}\left(2-x\right)\\ {\mathrm{log}}_{2}{5}^{2}={\mathrm{log}}_{2}\left(2-x\right)\\ 25=2-x\\ x=-23\end{array}$