Question 14 Marks
Find the values of n in the following:
If $\frac{9^{\text{n}}\times3^2\times3^{\text{n}}-27^{\text{n}}}{(3^3)^5\times2^3}=\frac{1}{27},$ find the value of n
If $\frac{9^{\text{n}}\times3^2\times3^{\text{n}}-27^{\text{n}}}{(3^3)^5\times2^3}=\frac{1}{27},$ find the value of n
Answer
View full question & answer→$\frac{9^{\text{n}}\times3^2\times3^{\text{n}}-27^{\text{n}}}{(3^3)^5\times2^3}=\frac{1}{27}$
$=\frac{(3^2)^\text{n}\times3^3\times3^{\text{n}}-(3^3)^{\text{n}}}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{(2\text{n}+2+\text{n})}-3^{3\text{n}}}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}\times3^2-3^{3\text{n}}}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}(3^2-1)}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}(9-1)}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}(8)}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}}{3^{15}}=\frac{1}{27}$
$=3^{3\text{n}-15}=\frac{1}{27}$
On equating the coefficient
$3\text{n} = 15 = -3$
$3\text{n} = -3+15$
$3\text{n}=12$
$\text{n}=4$
$=\frac{(3^2)^\text{n}\times3^3\times3^{\text{n}}-(3^3)^{\text{n}}}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{(2\text{n}+2+\text{n})}-3^{3\text{n}}}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}\times3^2-3^{3\text{n}}}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}(3^2-1)}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}(9-1)}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}(8)}{3^{15}\times2^3}=\frac{1}{27}$
$=\frac{3^{3\text{n}}}{3^{15}}=\frac{1}{27}$
$=3^{3\text{n}-15}=\frac{1}{27}$
On equating the coefficient
$3\text{n} = 15 = -3$
$3\text{n} = -3+15$
$3\text{n}=12$
$\text{n}=4$