Question 1204250




Let {{{x }}}be a real number such that {{{625^x=64 }}}

Then {{{125^x=a*sqrt(b)}}} What are {{{a }}}and {{{b}}}?


We have the equation {{{625^x = 64}}}

To simplify the equation, we can express both sides with the same base. We know that {{{625}}} can be expressed as {{{5^4}}} and {{{64}}} can be expressed as {{{2^6}}}.

Rewriting the equation, we have 

{{{(5^4)^x = 2^6}}}

Using the property of exponents, we can simplify further:

{{{5^(4x) = 2^6}}}

To find {{{x}}}, we need to equate the exponents:

{{{4x = 6}}}


Now, solving for{{{ x}}}:

{{{x = 6/4}}}


Now, we can calculate the value of {{{125^x}}} using the value of {{{x }}}we found:

{{{125^(6/4)=125^(3/2)}}}

since {{{125=5^3}}}, we have {{{(5^3)^(3/2)=5^(3(3/2))=5^(9/2)}}}

then

{{{125^x = sqrt(5^9)}}}

since {{{sqrt(5^9)=sqrt((5^4)^2*5)=5^4*sqrt(5)=625sqrt(5)}}}, then

{{{125^x = 625sqrt(5)}}}

so, comparing to {{{125^x=a*sqrt(b)}}}, we have

{{{a=625}}}

{{{b=5}}}