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# P2 - Pure maths - Logs continued watch

1. Find the non-zero values of x and y for which x^y = y^(3x) and y^3 = X^2

2. x^y=y^(3x)
ylogx=3xlogy
y/logy=3x/logx

y^3=x^2
3logy=2logx

You can then substitute one into the other so

y/logy=3x/(1.5logy)

so
3ylogy/2logy=3x

so 1.5y=3x

y/(2/3)logx=3x/logx
y=6xlogx/3logx
y=2x

So y=2x for all non-zero values of y and x.
Find the non-zero values of x and y for which x^y = y^(3x) and y^3 = X^2

x^y = y^(3x) -> y logx = 3x log y

y^3 = x^2 -> y = x^(2/3)

x^(2/3) log x = 3 x log [x^(2/3)] = 2 x log x

So logx = 0 and x = 1 and y = 1 is a possibility

or

x^(2/3) = 2 x

1/2 = x^(1/3)

1/8 = x and y = 1/4
4. How did I go wrong? Hm. Ah well, sorry.

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