Question

Question 99(iii)

Find the value of x, so that:
$(2^{-1}+4^{-1}+6^{-1}+8^{-1}{)}^x=1$

Answer 1

we have $(2^{-1}+4^{-1}+6^{-1}+8^{-1}{)}^x=1$
Using law of exponents $a^{-m}=\frac{1}{a^m}$
Then ${(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8})}^x=1\Rightarrow {\left(\frac{12+6+4+3}{24}\right)}^x=1\Rightarrow {\left(\frac{25}{24}\right)}^x=1$
this can be possible only if $x=0$, Since$a^0=1$