January 21 Information

a=5^x , b=5^y and a^y*b^x=25 Prove that xy=1. — Preceding unsigned comment added by Sahil shrestha ( talk • contribs) 09:42, 21 January 2017 (UTC) reply

We don't do homework for you, but have you tried taking logs to the base 5 for all three equations? Dbfirs 09:59, 21 January 2017 (UTC) reply

Expanding the third equation gives (5^x)^y * (5^y)^x = 25. Does that help? -- SGBailey ( talk) 14:04, 22 January 2017 (UTC) reply

You might also find the identity ${\displaystyle (b^{x})^{y}=b^{(xy)}}$ helpful. Smurrayinchester 16:39, 22 January 2017 (UTC) reply

And don't forget that ${\displaystyle b^{yx}=b^{xy}}$. StuRat ( talk) 17:44, 22 January 2017 (UTC) reply

Please see the image. Thank you. [1] 69.22.242.15 ( talk) 19:10, 21 January 2017 (UTC) reply

• Convenient rewriting: the first line reads: ${\displaystyle z=v\times f(u^{2}-v^{2})}$, prove ${\displaystyle v{\frac {\partial z}{\partial u}}+u{\frac {\partial z}{\partial v}}={\frac {uz}{v}}}$.

It seems that ${\displaystyle f}$ is a function of a single variable, so the " partial derivatives" ${\displaystyle {\frac {\partial f}{\partial u}},{\frac {\partial f}{\partial v}}}$ do not make any sense. Replacing them by ${\displaystyle f'(u^{2}-v^{2})}$, "the" derivative of f, corrects the proof. Tigraan ^{Click here to contact me} 21:13, 21 January 2017 (UTC) reply

Silly me. Thank you. 69.22.242.15 ( talk) 21:30, 21 January 2017 (UTC) reply

To crack the corrosion on some of my mental functions I decided to go through this explicitly:

• ${\displaystyle {\frac {\partial z}{\partial u}}={\frac {\partial (vf(u^{2}-v^{2}))}{\partial u}}=v{\frac {\partial f(u^{2}-v^{2})}{\partial u}}=v{\frac {\partial f(u^{2}-v^{2})}{\partial (u^{2}-v^{2})}}{\frac {\partial (u^{2}-v^{2})}{\partial u}}=2uvf'(u^{2}-v^{2})}$
• ${\displaystyle {\frac {\partial z}{\partial v}}={\frac {\partial v}{\partial v}}f(u^{2}-v^{2})+v{\frac {\partial f(u^{2}-v^{2})}{\partial v}}=f(u^{2}-v^{2})+v{\frac {\partial f(u^{2}-v^{2})}{\partial (u^{2}-v^{2})}}{\frac {\partial (u^{2}-v^{2})}{\partial v}}=f(u^{2}-v^{2})-2v^{2}f'(u^{2}-v^{2})}$
• So ${\displaystyle v{\frac {\partial z}{\partial u}}+u{\frac {\partial z}{\partial v}}=2uv^{2}f'(u^{2}-v^{2})+uf(u^{2}-v^{2})-2uv^{2}f'(u^{2}-v^{2})=uf(u^{2}-v^{2})={\frac {uz}{v}}}$

I don't guarantee I didn't do anything illegitimate ... it's been a while! Wnt ( talk) 17:59, 22 January 2017 (UTC) reply

What did you use to format the equations? How much more time would you guesstimate it took vs. handwriting? 69.22.242.15 ( talk) 22:28, 23 January 2017 (UTC) reply

My fingers, typing. It took me two edits, so you can sort of guesstimate the time from the history: about nine minutes for the larger part of it, I hate to say. [2] But I haven't used that format much. Wnt ( talk) 23:39, 23 January 2017 (UTC) reply

Wnt used Wikipedia's system for displaying mathematical equations, which itself is based on LaTeX.

LaTeX is the ubiquitous platform for creating academic articles and whitepapers in mathematical fields, and its format for creating equations is used on Wikipedia, math.stackexchange, Quora and elsewhere. You would do well to learn it.

On Wikipedia, a LaTeX formula is created by enclosing it with <math> </math> tags. You can also see Wnt's source code when you're in the dialog for editing this section. -- Meni Rosenfeld ( talk) 10:58, 24 January 2017 (UTC) reply