Question

Simplify and express the result in power notation with positive exponent.
(i) $(-4{)}^5÷(-4{)}^8$
(Ii) ${\left(\frac{1}{2^3}\right)}^2$
(iii) $(-3{)}^4\times {\left(\frac{5}{3}\right)}^4$
(iv) $(3^{-7}÷3^{-10})\times 3^{-5}$
(v) $2^{-3}\times (-7{)}^{-3}$

Answer 1

$\left(i\right)(-4{)}^5÷(-4{)}^8$

$=\frac{(-4{)}^5}{(-4{)}^8}=(-4{)}^{5-8}$

$=(-4{)}^{-3}=((-4{)}^3{)}^{-1}$

$=(-64{)}^{-1}$

$=\frac{-1}{64}$

$=\frac{-1}{2^6}$

$\left(ii\right){\left(\frac{1}{2^3}\right)}^2$

$={\left(\frac{1}{8}\right)}^2$

$=\frac{{\left(1\right)}^2}{8^2}$

$=\frac{1}{64}$

$=\frac{1}{2^6}$

$\left(iii\right)(-3{)}^4\times {\left(\frac{5}{3}\right)}^4$

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

$=\frac{3^4\times 5^4}{3^4}$

$=5^4$

$\left(iv\right)(3^{-7}÷3^{-10})\times 3^{-5}$

$=\frac{3^{-7}}{3^{-10}}\times 3^{-5}$

$=3^{-7-5+10}$

$=3^{-2}$

$=\frac{1}{3^2}$

$\left(v\right)2^{-3}\times (-7{)}^{-3}$

$=\frac{1}{2^3}\times \frac{-1}{(7{)}^3}$

$=\frac{-1}{(14{)}^3}$