Calculation

[Primarygun]

[MATH] log(x+y)/logx = log(x+y)/logy[/MATH]

 

Can anybody show a calculation?

I think that's a trick.

[Dave]

Sorry, what do you actually want to do with that equation?

[Primarygun]

solve for x in terms of y

[Dave]

Well, that looks pretty trivial from the face of it, but there are some technicalities. I don't know whether that's a typo, but you have the same numerator on both sides, so just cancel that out (unless x+y=1 in which case you're screwed) and then anti-log both sides.

[Freeman]

I got log y is positive or negative x....

[Dave]

y can't be negative x otherwise the equation wouldn't hold (log(0) not defined).

[Freeman]

Touché!

[Dave]

Cheers I got the solution y=x also.

 

I could swear that the question has a typo in it somewhere, that just seems too simple.

[Primarygun]

Well, that looks pretty trivial from the face of it, but there are some technicalities. I don't know whether that's a typo, but you have the same numerator on both sides, so just cancel that out (unless x+y=1 in which case you're screwed) and then anti-log both sides.

x+y=1

x=y

 

Ya there are two answers.

But what makes you know that x+y=1 is also possible.

Normally, don't us only cancel the log sign to find out x=y instead of minus the left side by the right side and then fing x+y=1?

[pulkit]

Look at the solution :

 

[MATH]\frac{\log(x+y)}{\log(x)} = \frac{\log(x+y)}{\log(y)}[/MATH]

[MATH]\log(x+y) \times (\frac{1}{\log(x)} - \frac{1}{\log(y)}) = 0[/MATH]

[MATH]\Rightarrow \log(x+y) = 0 \ldots or \ldots \frac{1}{\log(x)}=\frac{1}{\log(y)}[/MATH]

[MATH]\Rightarrow x+y=1 \ldots or \ldots \log(x)=\log(y)[/MATH]

[MATH]\Rightarrow x+y=1 \ldots or \ldots x=y [/MATH]

[Primarygun]

yes I know it.

But why do you use this method?

Normally, don't we use the cross-method calculation only?

If I don't say there has a trick or there are two possible answers, how many answers do you desire to get?

[pulkit]

When you cancel out a term from both sides, you always assume that this term is non-zero. Hence you must consider the case of this term being zero seperately. In the particular question this term is [MATH]\log(x+y)[/MATH] and [MATH]x+y=1[/MATH] comes out of the case when it is zero.

[Dave]

sigh. I suck

[Primarygun]

When you cancel out a term from both sides, you always assume that this term is non-zero. Hence you must consider the case of this term being zero seperately. In the particular question this term is [MATH]\log(x+y)[/MATH] and [MATH]x+y=1[/MATH'] comes out of the case when it is zero.

Thanks.

Is there any other cases for other regions of mathematics?

[pulkit]

Is there any other cases for other regions of mathematics?

What do you mean ?

[Primarygun]

such as normal algebra.