# Math Help - Exponents (x and 2x)??

1. ## Exponents (x and 2x)??

I can't solve this for the life of me; help ?

2.71 = (1.0915)^(-x) + 3(1.0915)^(-2x)
How do we solve such equations? I can't factor because I'm left with an x in my exponent ... I'm stumped.
Is there even a solution?

2. Did you try?

$2.71 = 1.0915^{-x} + 3(1.0915)^{-2x}$

$2.71 = 1.0915^{-x} + 3((1.0915)^{-x}\times (1.0915)^{2})$

Oops, sorry that ain't right!

3. I honestly did and I can't figure out any way to isolate x. Am I missing something plainly obvious here??
I tried to factor out (1.0915)^x, but that didn't solve the problem at all.

Isn't that equal to (1.0915)^(2-x) however? Exponents add??

4. Yep that was a boo boo on my behalf.

Might have to solve this one numerically. I'll have a think about it.

5. Thank you very much!
(What does numerically mean??)
I'm in my first university year just to let you know - it's possible that I perhaps got the equation wrong from the problem too. So don't go to too much trouble, just wondered if I was missing something obvious!

6. I have found a solution numerically $\approx 2.575$

7. Originally Posted by Volcanicrain
Thank you very much!
(What does numerically mean??)
I just put the equation in an excel sheet and tried a list of different numbers.

8. Thank, that does work.
Now I'll figure out how to prove it. Thanks =)

9. Originally Posted by Volcanicrain
2.71 = (1.0915)^(-x) + 3(1.0915)^(-2x)
Make your life easier: let k = 1.0915; then you have:
k^(-x) + 3k^(-2x) = 2.71
Get rid of minuses:
1/k^x + 3/k^(2x) = 2.71
Multiply by k^(2x):
k^x + 3 = 2.71k^(2x)
Rearrange;
2.71(k^x)^2 - k^x - 3 = 0

You now have a quadratic...can you finish it?

You may find it easier to let a = k^x; then:
2.71a^2 - a - 3 = 0 : get my drift?
Solve for a; then 1.0915^x = a ; x = log(a) / log(1.0915)

10. Originally Posted by Wilmer
Make your life easier: let k = 1.0915; then you have:
k^(-x) + 3k^(-2x) = 2.71

You now have a quadratic...can you finish it?
*kicks myself*

11. Wow thank you Wilmer!
I was playing around with the idea of a quadratic... I was trying to complete a square or something but had not thought of that.
It makes perfect sense to remove the negative exponents by multiplying by k^2x all across.