Question

Find the value of 'n' in each of the following :
i) ${\left(\frac{2}{3}\right)}^3\times {\left(\frac{2}{3}\right)}^5={\left(\frac{2}{3}\right)}^{n-2}$
ii) $(-3{)}^{n+1}\times (-3{)}^5=(-3{)}^{-4}$
iii) $7^{2n+1}÷49=7^3$

Answer 1

(i) ${\left(\frac{2}{3}\right)}^3\times {\left(\frac{2}{3}\right)}^5={\left(\frac{2}{3}\right)}^{n-2}$

$\Rightarrow (n-2)=5+3\Rightarrow n=10$

(ii) $(-3{)}^{n+1}x(-3{)}^5=\left)\right(-3{)}^{-4}$

$\Rightarrow n+1+5=-4$

$\Rightarrow n=-10$

(iii) $7^{2n+1}÷49=7^3$

$\Rightarrow 7^{2n+1}÷7^2=7^3$

$\Rightarrow (2n+1)-2=3$

$\Rightarrow 2n=4$

$n=2$