Did you try?
Oops, sorry that ain't right!
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!
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)
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.
I am really grateful for your answer!! Thank you so much!