Question

Find the value of $x$
i) ${\left(\frac{4}{3}\right)}^{-4}\times {\left(\frac{4}{3}\right)}^{-5}={\left(\frac{4}{3}\right)}^{-3x}$
ii) $7^x÷7^3=7^5$
iii) $4^{2x+1}÷16=64$

Answer 1

(i) $(\frac{4}{3}{)}^{-4}\times (\frac{4}{3}{)}^{-5}=(\frac{4}{3}{)}^{-3x}$

$⟹(\frac{4}{3}{)}^{-4+5}=(\frac{4}{3}{)}^{-3x}$ (since, $a^m\times a^n=a^{m+n}$)

$⟹\left(\frac{4}{3}\right)=(\frac{4}{3}{)}^{-3x}$

bases are equal then equate the powers.

$⟹1=-3x$

$⟹x=\frac{-1}{3}$

(ii) $\frac{7^x}{7^3}=7^5$

$⟹7^{x-3}=7^5$ (since, $a^m/a^n=a^{m-n}$)

bases are equal then equate the powers.

$⟹x-3=5$

$⟹x=8$

(ii) $\frac{4^{2x+1}}{4^2}=4^3$

$⟹4^{2x+1-2}=4^3$ (since, $a^m/a^n=a^{m-n}$)

bases are equal then equate the powers.

$⟹2x-1=3$

$⟹2x=4$

$⟹x=2$