Algebra Problem of the Day

[BigBeachBanana]

Prove that [imath]a^x-b^y=0[/imath], where [imath]x=\sqrt{\log_ab},\,y = \sqrt{\log_{b}a}[/imath] ; [imath]a,b >0\,\, \& \,\, a,b \neq 1.[/imath]

 

[Steven G]

Enough to show that a^x=b^y

x= sqrt(log_ab) and y=sqrt(log_ba)

So xy = 1 or y=1/x = 1/sqrt(log_ab)

(a^x)^x= a^x2

(b^y)^x=b^xy=b

Does a^x2 = b?

IFF log_ba^x2 = 1

log_ba^x2 = x² log_ba = log_ab * log_ba = 1

 

[Cubist]

I managed to follow @Steven G 's solution, but I'd have used the word "therefore" instead of "or" here...

So xy = 1 or y=1/x = 1/sqrt(log_ab)

...since this line confused me for a ridiculous amount of time

Here's my (probably slow) approach...
[math] a^{\sqrt{\log_{a}{b}}} = b^{\sqrt{\log_{b}{a}}} [/math][math] \ln\left(a^{\sqrt{\log_{a}{b}}}\right) = \ln \left(b^{\sqrt{\log_{b}{a}}}\right) [/math][math] \sqrt{ \log_{a}{b}} \cdot \ln a - \sqrt{\log_{b}{a}} \cdot \ln b= 0[/math]

---

[math] \text{L.H.S} = \sqrt{ \log_{a}{b}} \cdot \ln a - \sqrt{\log_{b}{a}} \cdot \ln b [/math][math] = \sqrt{\frac{\ln b}{\ln a}} \cdot \ln a - \sqrt{ \frac{\ln a}{\ln b}} \cdot \ln b[/math][math]= \sqrt{\ln b \cdot \ln a} - \sqrt{ \ln a \cdot \ln b } [/math][math]=0 = \text{R.H.S} [/math]

 

[BigBeachBanana]

Yes the fast way is to notice [imath]y=\frac{1}{x} \implies a^x=b^{\frac{1}{x}}\implies a^{x^2}=b[/imath], where [imath]x^2=\log_a{b}[/imath].
It follows [imath]a^{\log_a{b}}=b.[/imath]

 

[Steven G]

I managed to follow @Steven G 's solution, but I'd have used the word "therefore" instead of "or" here...

If you keep using those big words you're going to get yourself in trouble!

 

[Cubist]

If you keep using those big words you're going to get yourself in trouble!

Therefore a continued violation would necessitate that my displacement from the interception of two orthogonal planes (both possessing a normal with negligible z components) ought to be reduced in magnitude?

Spoiler

So if I keep on, then I'm off to the corner?