Find x, if

Solution:

(i) We have:

$\left(\frac{1}{4}\right)^{-4} \times\left(\frac{1}{4}\right)^{-8}=\left(\frac{1}{4}\right)^{-4 x}$

$\left(\frac{1}{4}\right)^{-12}=\left(\frac{1}{4}\right)^{-4 x} \quad\left(a^{m} \times a^{n}=a^{m+n}\right)$

$-12=-4 x$

$3=x$

$x=3$

(ii) We have:

$\left(\frac{-1}{2}\right)^{-19} \times\left(\frac{-1}{2}\right)^{8}=\left(\frac{-1}{2}\right)^{-2 x+1}$

$\left(\frac{-1}{2}\right)^{-11}=\left(\frac{-1}{2}\right)^{-2 x+1} \quad\left(a^{m} \times a^{n}=a^{m+n}\right)$

$-11=-2 x+1$

$-12=-2 x$

$6=x$

x = 6

(iii) We have:

$\left(\frac{3}{2}\right)^{-3} \times\left(\frac{3}{2}\right)^{5}=\left(\frac{3}{2}\right)^{2 x+1}$

$\left(\frac{3}{2}\right)^{2}=\left(\frac{3}{2}\right)^{2 x+1}$

$2=2 x+1$

$1=2 x$

$\frac{1}{2}=x$

x = 1/2

(iv) We have:

$\left(\frac{2}{5}\right)^{-3} \times\left(\frac{2}{5}\right)^{15}=\left(\frac{2}{5}\right)^{2+3 x}$

$\left(\frac{2}{5}\right)^{12}=\left(\frac{2}{5}\right)^{2+3 x}$

$12=2+3 x$

$10=3 x$

$\frac{10}{3}=x$

x = 10/3

(v) We have:

$\left(\frac{5}{4}\right)^{-x} \div\left(\frac{5}{4}\right)^{-4}=\left(\frac{5}{4}\right)^{5}$

$\left(\frac{5}{4}\right)^{-x+4}=\left(\frac{5}{4}\right)^{5}$

$-x+4=5$

$-x=1$

$x=-1$

x = −1

(vi) We have:

$\left(\frac{8}{3}\right)^{2 x+1} \times\left(\frac{8}{3}\right)^{5}=\left(\frac{8}{3}\right)^{x+2}$

$\left(\frac{8}{3}\right)^{2 x+6}=\left(\frac{8}{3}\right)^{x+2}$

$2 x+6=x+2$

$x=-4$

x = −4