(ask) Proving Logarithmic Equation

J

Jglow

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Type: Proving, Topic: Logarithmic Equation


Question: If [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH] [math]= 1[/math], prove [MATH]\frac{1}{\log_{x+y} a } [/MATH] [math]=[/math] [MATH]\frac{1}{\log_x a }[/MATH] [math]+[/math] [MATH]\frac{1}{\log_x y }[/MATH]

My attempt: trying to prove from the right side of the equation.

[MATH]\frac{1}{\log_x a }[/MATH] [math]+[/math] [MATH]\frac{1}{\log_x y }[/MATH]
[math]= \frac{1}{\frac{\log_a a }{\log_a x }} + \frac{1}{\frac{\log_y y }{\log_y x }}[/math]
[math]= \log_a x + \log_a y[/math]
[math]= \log_a {xy}[/math]
[math]= \frac{\log_{x+y} xy }{\log_{x+y} a }[/math]
[math]= \frac{\log_{x+y} x + \log_{x+y} y}{\log_{x+y} a }[/math]
[math]= \frac{\frac{\log_x x}{\log_x {x+y}} + \frac{log_y y}{\log_y {x+y}}}{\log_{x+y} a }[/math]
[math]= \frac{\frac{1}{\log_x {x+y}} + \frac{1}{\log_y {x+y}}}{\log_{x+y} a }[/math]

Then I was stuck, having no idea to continue solving it...
My idea is trying to make the numerator [math]{\frac{1}{\log_x {x+y}} + \frac{1}{\log_y {x+y}}}[/math] to [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH], so that I could make it [math]= 1[/math], as the question has given [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH] [math]= 1[/math], but I can't think of any other ways to make the numerator to [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH].
 
For finding a proof, it is frequently clever to assume as true what is to be proved and work backward to a known truth. That of course is not a proof, but it may indicate a proof in reverse. NEVERTHELESS, it is always best to try FIRST from what you KNOW is true or given.

[MATH]1 = \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x + y}{xy} \implies xy = x + y.[/MATH]
[MATH]\therefore log_{x + y}(a) = log_{xy}(a) = \dfrac{log_x(a)}{log_x(xy)} = \dfrac{log_x(a)}{log_x(x) + log_x(y)} = \dfrac{log_x(a)}{1+ log_x(y)}.[/MATH]
[MATH]\therefore \dfrac{1}{log_{x+y}(a)} - \dfrac{1}{log_x(a)} = \dfrac{1+ log_x(y)}{log_x(a)} - \dfrac{1}{log_x(a)} = \dfrac{log_x(y)}{log_x(a)} \implies[/MATH]
[MATH]\dfrac{1}{log_{x+y}(a)} = \dfrac{1}{log_x(a)} + \dfrac{log_x(y)}{log_x(a)}. \text {Q.E.D.}[/MATH]
 
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Without wishing to be a nuisance, unless I am missing some information, the statement in the question is not true and so can't be proved.
Also, (therefore) what has been proved above is not the statement in the question (true, but not Q.E.D.).

Jglow said:
Question: If [MATH]\frac{1}{x} + \frac{1}{y} =1,[/MATH] prove [MATH]\frac{1}{\log_{x+y} a } = \frac{1}{\log_x a } + \frac{1}{\log_x y } [/MATH]
Click to expand...

JeffM said:
[MATH]\dfrac{1}{log_{x+y}(a)} = \dfrac{1}{log_x(a)} + \dfrac{log_x(y)}{log_x(a)}. \text {Q.E.D.}[/MATH]
Click to expand...
 
JeffM said:
[MATH]\dfrac{1}{log_{x+y}(a)} = \dfrac{1}{log_x(a)} + \dfrac{log_x(y)}{log_x(a)}. \text {Q.E.D.}[/MATH]
Click to expand...
I think @JeffM is correct, and the above equals the following that looks similar to the original proof (but not quite the same)...

[MATH]= \dfrac{1}{log_x(a)} + \dfrac{1}{log_y(a)} [/MATH]

Therefore I suspect a typo in @lex 's original post since the identity quoted isn't true. Try plugging in some numbers, say x=5, y=5/4, a=7

[MATH]\frac{1}{\log_{x+y} a } [/MATH] [math]=[/math] [MATH]\frac{1}{\log_x a }[/MATH] [math]+[/math] [MATH]\frac{1}{\log_x y } \,[/MATH] ...this is the original eqn, probably with a typo

0.94176 = 8.03965, hmmm ?

EDIT: The numbers work in JeffM's version
 
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lex said:
Also, (therefore) what has been proved above is not the statement in the question (true, but not Q.E.D.).
Click to expand...


Yes @JeffM's work is correct - as I say 'true, if not quod erat demonstrandum (what was to be demonstrated)!

You're absolutely right of course - the question should have been:

Question: If [MATH]\frac{1}{x} + \frac{1}{y} =1,[/MATH] prove [MATH]\frac{1}{\log_{x+y} a } = \frac{1}{\log_x a } + \frac{1}{\log_y a } [/MATH]
(which has the required symmetry too)
 
@lex

Oh, you are right. I gave a valid proof, but it was not of the proposition actually propounded. That’s funny.

I suspect, however, that it was the proposition intended.
 
Thank you all for spending time to read my post and answer my question. Thank you @JeffM for your solution and your idea of solving from what is true or given, I never thought of solving in that way.

For the typo of the question, I have checked what I have typed in my post with the source of the question. The question that I typed is exactly the same as the one in the book where the question from.

Also, I used a calculator to calculate the statement of the question to see whether both sides of the equation are equal by assuming the variables to some different numbers. As a result, both sides of the equation are not equal to each other. Perhaps the question should be as indicated by @lex :
lex said:
If [math]\displaystyle \frac{1}{x} + \frac{1}{y} =1[/math], prove [math]\frac{1}{\log_{x+y} a } = \frac{1}{\log_x a } + \frac{1}{\log_y a }[/math]
Click to expand...





As for the convenience of future readers, this is a summary of the question and solution:

Question:
lex said:
If [math]\displaystyle \frac{1}{x} + \frac{1}{y} =1[/math], prove [math]\frac{1}{\log_{x+y} a } = \frac{1}{\log_x a } + \frac{1}{\log_y a }[/math]
Click to expand...
(thanks @lex for the indication)

Solution:
refer to the first reply of this thread by @JeffM and continue your working from the last row of the solution by @JeffM to this by @Cubist :
Cubist said:
[math]\displaystyle = \dfrac{1}{log_x(a)} + \dfrac{1}{log_y(a)} [/math]
Click to expand...
(thanks @JeffM for your solution and also @Cubist for your extension to the solution by @JeffM )
 
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Sorry @lex I was getting you mixed up with @Jglow when I wrote my post#4, and I didn't read your post properly ! :oops:

@Jglow perhaps you can mark this thread as solved now?
 
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