Evaluating the value of exponential expression

Question

What is the value of:

$\frac{2a}{a^{x-y}-1}+\frac{2a}{a^{y-x}-1}$

I tried this:

$(\frac{a^{x-y}-1}{2a})^{-1}+(\frac{a^{y-x}-1}{2a})^{-1}$

$\frac{({a^{x-y}-1})^{-1}+({a^{y-x}-1})^{-1}}{2a^{-1}}$

But then I was stuck... any ideas?

what do you mean with "value" only variables are given – Dr. Sonnhard Graubner Oct 21 '14 at 06:28 — Dr. Sonnhard Graubner, Oct 21 '14 at 06:28
The answer can be “a” or “-3a”, not only numbers. – Arda Arman Oct 21 '14 at 06:32 — Arda Arman, Oct 21 '14 at 06:32

score 1 · Accepted Answer · answered Oct 21 '14 at 06:29

1

Hint: the second fraction can be written $$\frac{2a}{a^{y-x}-1}=\frac{(2a)a^{x-y}}{1-a^{x-y}}$$ and now you can collect your two fractions over a common denominator.

answered Oct 21 '14 at 06:29

David

82,662

score 1 · Answer 2 · answered Oct 21 '14 at 06:36

1

HINT:

$$\frac1{a^{x-y}-1}=\frac1{\dfrac{a^x}{a^y}-1}=\frac{a^y}{a^x-a^y}$$

answered Oct 21 '14 at 06:36

lab bhattacharjee

274,582

score 0 · Answer 3 · answered Oct 21 '14 at 06:33

0

we have $\left(\frac{1}{a^{x-y}-1}+\frac{1}{\frac{1}{a^{x-y}}-1}\right)$= $2a\left(\frac{1}{a^{x-y}-1}-\frac{a^{x-y}}{a^{x-y}-1}\right)=-2a$

answered Oct 21 '14 at 06:33

Dr. Sonnhard Graubner

95,283

yes the answer is $-2a$ a parameter? – Dr. Sonnhard Graubner Oct 21 '14 at 06:34

Evaluating the value of exponential expression

3 Answers3