Simplify logarithms

[petarantes]

Anybody help? I tried but couldn't solve.


If a²+b² = 1 simplify:

[MATH]T=\frac{\log(a+b+1)-\log(a-b+1)}{\log a-\log(b+1)} [/MATH]

Spoiler: Answer

-1

 

[ksdhart2]

You say "[you] tried," but what did you try? Presumably you started by reviewing the rules of logarithms, specifically noting that the difference of two logarithms is equal to the logarithm of the quotient of the individual components... but then what? Where did that lead you?

Please (re)read the Read Before Posting thread that's stickied at the top of each sub-forum and comply with the rules found therewithin. In particular, please share with us all of the work you've done on this problem, even the parts you know for sure are wrong. Thank you.

Edit: Whoops. I missed a very important word that totally changes the meaning of what I said. The inserted word is underlined.

 

[petarantes]

I try:
[MATH]T=\frac{log(\frac{a+b+1}{a-b+1})}{log(\frac{a}{b+1})}= [/MATH]
(a+b)(a-b) = a²-b²

but I don't see how to substitute this

 

[Dr.Peterson]

There may be a quicker way, but the obvious thing you can do is to replace a^2 with 1 - b^2 everywhere and simplify each part. This does lead to the answer (I think).

Edit: Of course, this requires some conditions, such as that a > 0, and maybe more. But I made the replacement and graphed the unsimplified expression, and saw the expected graph come out, an interrupted horizontal line at y = -1. Then I tried doing the simplification, and unless I was too hasty, it did work.

 

[ksdhart2]

Actually, I think I disagree with Dr. Peterson here. I believe that the answer of T = -1 is wholly incorrect. In fact, I posit that there is no way to determine a numerical solution (though I'm certainly open to a correction if I'm wrong on this).

 

[petarantes]

My colleague fantacelle managed to solve; Follow the resolution
[MATH] T=\frac{log(\frac{a+b+1}{a-b+1})}{log(\frac{a}{b+1})}=log_{\frac{a}{b+1}}(\frac{a+b+1}{a-b+1}) \\ \therefore \frac{a^T}{(b+1)^T}=\frac{a+b+1}{a-b+1} \\ a^2=1-b^2 \rightarrow \frac{a^2}{b+1}=1-b \rightarrow \frac{a}{b+1}=\frac{1-b}{a} \\ \frac{a+(b+1)}{b+1}=\frac{1-b+(a)}{a} \rightarrow \frac{a+b+1}{a-b+1}=\frac{b+1}{a} \\ (\frac{a}{b+1})^T=\frac{b+1}{a}\\ \therefore T=-1 [/MATH]

 

[Dr.Peterson]

Very nice. That's a direction I had considered, but left for you to try!

Of course, there are a number of conditions that have been left unstated, which in summary amount to a and b being real and non-zero.

 

[MarkFL]

Not nearly as elegant an approach, but here it is nonetheless.

From the given constraint, we have:

[MATH]-b+1=\frac{a^2}{b+1}[/MATH]
[MATH]b+1=\frac{a^2}{-b+1}[/MATH]
Hence:

[MATH]T=\frac{\log\left(a+\dfrac{a^2}{-b+1}\right)-\log\left(a+\dfrac{a^2}{b+1}\right)}{\log(a)-\log(b+1)}[/MATH]
[MATH]T=\frac{\log\left(1+\dfrac{a}{-b+1}\right)-\log\left(1+\dfrac{a}{b+1}\right)}{\log(a)-\log(b+1)}[/MATH]
[MATH]T=\frac{\log\left(\dfrac{a-b+1}{-b+1}\right)-\log\left(\dfrac{a+b+1}{b+1}\right)}{\log(a)-\log(b+1)}[/MATH]
[MATH]T=\frac{\log(a-b+1)-\log(-b+1)-\log(a+b+1)+\log(b+1)}{\log(a)-\log(b+1)}[/MATH]
[MATH]T=\frac{-(\log(a+b+1)-\log(a-b+1))-\log(-b+1)+\log(b+1)}{\log(a)-\log(b+1)}[/MATH]
[MATH]2T=\frac{\log(b+1)-\log(-b+1)}{\log(a)-\log(b+1)}[/MATH]
[MATH]2T=\frac{\log(b+1)-\log\left(\frac{a^2}{b+1}\right)}{\log(a)-\log(b+1)}[/MATH]
[MATH]2T=\frac{2\log(b+1)-2\log(a)}{\log(a)-\log(b+1)}=-2\implies T=-1[/MATH]

 

[petarantes]

grateful to everyone for their help