**5.6 Indices and Logarithms, SPM Practice (Long Questions)**

Question 1:

Question 1:

(a) Find the value of

i. 2 log

_{2}12 + 3 log_{2}5 – log_{2}15 – log_{2}150. ii. log

_{8}32(b) Shows that 5

^{n}^{ }+ 5^{n}^{ +}^{ }^{1 }+ 5^{n}^{ +}^{ }^{2 }can be divided by 31 for all the values of*n*which are positive integer.

*Solution:***(a)(i)**

2 log

_{2}12 + 3 log_{2}5 – log_{2}15 – log_{2}150= log

_{2}12^{2}+ log_{2}5^{3}– log_{2}15 – log_{2}150
$={\mathrm{log}}_{2}\frac{{12}^{2}\times {5}^{3}}{15\times 150}$

= log

_{2}8= log

_{2}2^{3}= 3

**(a)(ii)**

$\begin{array}{l}{\mathrm{log}}_{8}32=\frac{{\mathrm{log}}_{2}32}{{\mathrm{log}}_{2}8}\\ \text{}=\frac{{\mathrm{log}}_{2}{2}^{5}}{{\mathrm{log}}_{2}{2}^{3}}=\frac{5}{3}\end{array}$

**(b)**

5

^{n}^{ }+ 5^{n}^{ +}^{ }^{1 }+ 5^{n}^{ +}^{ }^{2}= 5

^{n}^{ }+ (5 × 5^{n}^{ }) + (5^{2}× 5^{n}^{ })= 5

^{n}^{ }(1 + 5 + 5^{2})= 31 × 5

^{n}^{ }Therefore, 5

^{n}^{ }+ 5^{n}^{ +}^{ }^{1 }+ 5^{n}^{ +}^{ }^{2 }can be divided by 31 for all the values of*n*which are positive integer.**Question 2:**

(a) Given log

_{10}*x*= 3 and log_{10}*y*= –2. Shows that 2*xy*– 10000*y*^{2}= 19.(b) Solve the equation log

_{3}*x*= log_{9}(*x*+ 6).

*Solution:***(a)**

log

_{10}*x*= 3 → (*x*= 10^{3})log

_{10}*y*= –2 → (*y*= 10^{-2})2

*xy*– 10000*y*^{2}= 19Left hand side:

2

*xy*– 10000*y*^{2}= 2 × 10

^{3 }× 10^{-2 }– 10000 (10^{-2})^{2}= 20 – 10000 (10

^{-4})^{}= 20 – 1

= 19

= right hand side

**(b)**

$\begin{array}{l}{\mathrm{log}}_{3}x={\mathrm{log}}_{9}\left(x+6\right)\\ {\mathrm{log}}_{3}x=\frac{{\mathrm{log}}_{3}\left(x+6\right)}{{\mathrm{log}}_{3}9}\\ {\mathrm{log}}_{3}x=\frac{{\mathrm{log}}_{3}\left(x+6\right)}{{\mathrm{log}}_{3}{3}^{2}}\\ {\mathrm{log}}_{3}x=\frac{{\mathrm{log}}_{3}\left(x+6\right)}{2}\end{array}$

2log

_{3}*x*= log_{3}(*x*+ 6)log

_{3}*x*^{2}= log_{3}(*x*+ 6)*x*

^{2}=

*x*+ 6

*x*

^{2}–

*x*– 6 = 0

(

*x*+ 2) (*x*– 3) = 0*x =*– 2 atau 3.

log

Jadi, _{3 }(– 2) not accepted (logarithm of a negative number is undefined)*x =*3.