Prove that if x^y = y^x, then ln(x)/x = ln(y)/y.

justaguide | College Teacher | (Level 2) Distinguished Educator

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It is given that x^y = y^x

Take the logarithm to the base e, or ln, of both the sides

=> ln [ x^y] = ln [ y^x]

Use the property of log that log a^b = b*log a

=> y*ln x = x*ln y

divide both the sides by xy

=> [y*ln x]/xy = [x*ln y]/xy

=> (ln x)/x = (ln y)/y

If x^y = y^x, then this proves that (ln x)/x = (ln y)/y

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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It is given that x^y = y^x.

Take the natural logarithm of both the sides:

ln(x^y) = ln(y^x)

Use the property of logarithm ln a^x = x*ln a

y*ln x = x*ln y

Divide both sides by x*y

(y*ln x)/(x*y) = (x*ln y)/x*y)

ln x/x = ln y/y

This proves that ln x/x = ln y/y if x^y = y^x.

bhalachandra | College Teacher | (Level 1) eNoter

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x ^ y  =  y ^ x            ...                       (Given)

Taking  log  to  the  base  e  on  both  the  sides, we  get

y* ln (x)  =  x * ln (y)            [as  ln (a ^ b)  =  b*ln(a)]

Hence,  ln(x) /x   =  ln(y) / y