Question 763181
Show that {{{log(9,(x/y))=(1/2)(log(3,x)-log(3,y))}}}
{{{log(3,x/y) = log(9,x/y)/log(3,9) = log(9,x/y)/2}}}
{{{log(b,x/y) = log(b,x) - log(b,y)}}} b = any base
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Hence, find the values of x and y which satisfies the equation 
{{{log(9,x/y)=1/4}}} and {{{3log(3,x)=2log(3,y)}}}
{{{log(9,x/y)=1/4}}} --> {{{log(3,x/y)=1/2}}}
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{{{log(3,x/y)=1/2}}}
{{{3log(3,x)=2log(3,y)}}} --> {{{log(3,x^3)=log(3,y^2)}}}
{{{x^3 = y^2}}}
{{{y = x^(1.5)}}} (fractional exponents don't display well)
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{{{log(3,x/y)=1/2}}}
{{{log(3,x/x^1.5)=1/2}}}
{{{log(3,x^(-0.5)) = 1/2}}}
{{{1/sqrt(x) = sqrt(3)}}}
x = 1/3
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y = x^1.5
{{{y = sqrt(3)/9}}}