document.write( "Question 37838: Prove that if x^y=y^x then (x/y)^(x/y)=x^((x/y)-1) \n" ); document.write( "

Algebra.Com's Answer #23387 by ilana(307) $\"\"$ $\"About$

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\n" ); document.write( "x^y=y^x, so x^y/y^x=1. We want it to look like x^(x/y)/y^(x/y), so first we can raise each side of our equation to 1/y. So x^y/y^x=1 becomes (x^y)^(1/y)/(y^x)^(1/y)=1^(1/y), so x/(y^(x/y))=1. Now we just need x^(x/y) in the numerator. If we multiply (a^1)*(a^x), we get a^(x+1). So to get from x^1 to x^(x/y), we use the equation 1+m(or any variable)=x/y, so m=x/y-1.
\n" ); document.write( "So finally, we multiply our equation by x^((x/y)-1), and:
\n" ); document.write( "x/(y^(x/y))=1 goes to x*x^((x/y)-1)/(y^(x/y))=1*x^((x/y)-1), so:
\n" ); document.write( "(x*(x/y))/(y^(x/y)) = x^((x/y)-1), which completes our proof:). \n" ); document.write( "

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