Question 762788
{{{(1/128)^(-1/7)=128^(1/7)=(2^7)^(1/7)=2^(7*(1/7))=2^1=highlight(2)}}}
 
If further explanations are needed:
 
The definition of power with a negative exponent for a positive base {{{a}}} and a real exponent {{{x}}} is as follows.
{{{a^(-x)=1/a^x}}}
So {{{a^(-1/7)=1/a^(1/7)=1^(1/7)/a^(1/7)=(1/a)^(1/7)}}}
If {{{a=1/128}}}, then {{{1/((1/128))=128}}} and
{{{(1/128)^(-1/7)=128^(1/7)}}}
 
{{{128=2^7}}} and one the properties of powers is that for a positive base {{{a}}} and real exponents {{{x}}} and {{{y}}},
{{{(a^x)^y=a^(xy)}}}
so if {{{a=128}}}, {{{x=7}}} and {{{y=1/7}}}, then
{{{128^(1/7)=(2^7)^(1/7)=2^(7*(1/7))=2^1=highlight(2)}}}