Question

Find the values of n in each of the following:
(i) $5^{2n}\times 5^3=5^{11}$
(ii) $9\times 3^n=3^7$
(iii) $8\times 2^{n+2}=32$
(iv) $7^{2n+1}÷49=7^3$
(v) ${\left(\frac{3}{2}\right)}^4\times {\left(\frac{3}{2}\right)}^5={\left(\frac{3}{2}\right)}^{2n+1}$
(vi) ${\left(\frac{2}{3}\right)}^{10}+{\left\{{\left(\frac{3}{2}\right)}^2\right\}}^5={\left(\frac{2}{5}\right)}^{2n-2}$

Answer 1

We have

(i) 5²ⁿ x 5³ = 5¹¹
= 5²ⁿ⁺³ = 5¹¹
On equating the coefficients, we get
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n =$\frac{8}{2}=4$
(ii) 9 x 3ⁿ = 3⁷
= (3)² x 3ⁿ = 3⁷
= (3)²⁺ⁿ = 3⁷
On equating the coefficients, we get
2 + n = 7
⇒ n = 7 - 2 = 5

(iii) 8 x 2ⁿ⁺² = 32
= (2)³ x 2ⁿ⁺² = (2)⁵ [since 2³ = 8 and 2⁵ = 32]
= (2)³⁺ⁿ⁺² = (2)⁵
On equating the coefficients, we get
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 -5
⇒ n = 0

(iv) 7²ⁿ⁺¹ ÷ 49 = 7³
= 7²ⁿ⁺¹ ÷ 7² = 7³ [since 49 = 7²]
$=\frac{7^{2n+1}}{7^2}=7^3$
$=7^{2n+1-2}=7^3[\mathrm{since}\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}]$
= 7^2n-1 =7³
On equating the coefficients, we get
2n - 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n = $\frac{4}{2}=2$
(v) ${\left(\frac{3}{2}\right)}^4\times {\left(\frac{3}{2}\right)}^5={\left(\frac{3}{2}\right)}^{2n+1}$
$={\left(\frac{3}{2}\right)}^{(4+5)}={\left(\frac{3}{2}\right)}^{(2n+1)}$
$={\left(\frac{3}{2}\right)}^9={\left(\frac{3}{2}\right)}^{2n+1}$
On equating the coefficients, we get
2n + 1 = 9
⇒ 2n = 9 - 1
⇒ 2n = 8
⇒ n =$\frac{8}{2}=4$
(vi) ${\left(\frac{2}{3}\right)}^{10}\times {\left\{{\left(\frac{3}{2}\right)}^2\right\}}^5={\left(\frac{2}{5}\right)}^{2n-2}$
$={\left(\frac{2}{3}\right)}^{10}\times {\left(\frac{3}{2}\right)}^{10}={\left(\frac{2}{5}\right)}^{2n-2}$
$=\frac{2^{10}\times 3^{10}}{3^{10}\times 2^{10}}={\left(\frac{2}{5}\right)}^{2n-2}$
$=1={\left(\frac{2}{5}\right)}^{2n-2}$
$={\left(\frac{2}{5}\right)}^0={\left(\frac{2}{5}\right)}^{2n-2}[\mathrm{since}{\left(\frac{2}{5}\right)}^0=1]$
On equating the coefficients, we get
⇒ 0 = 2n - 2
⇒ 2n = 2
⇒ n = $\frac{2}{2}=1$