Question

Question 177(i)

Find x.
$-{\left(\frac{1}{7}\right)}^{-5}÷-{\left(\frac{1}{7}\right)}^{-7}=(-7{)}^x$

Answer 1

${(-\frac{1}{7})}^{-5}÷{(-\frac{1}{7})}^{-7}=(-7{)}^x$

Using law of exponents, $a^m÷a^n=(a{)}^{m-n}and{a}^{-m}={\left(\frac{1}{a}\right)}^m$

$\therefore {(-\frac{1}{7})}^{-5}÷{(-\frac{1}{7})}^{-7}=(-7{)}^x$

$\therefore (-7{)}^{-5}÷(-7{)}^{-7}=(-7{)}^x$

$\Rightarrow (-7{)}^{-5-(-7)}=(-7{)}^x$

$\Rightarrow (-7{)}^{-5+7}=(-7{)}^x$

$\Rightarrow (-7{)}^2=(-7{)}^x$

Since, the base is same on both sides, we can equate the powers.
On equating the powers of (-7), we get,
x = 2