# Thread: Equation that makes my life harder

1. ## Equation that makes my life harder

Hello everybody.
I have an equation that I am trying to solve for 7 hours. But nothing made me success, unfortunately. So I decided to ask for help there.

I will be very lucky if somebody helps me to solve my problem.

P.S.: Sorry for my English, I am only beginner now.

2. ## Re: Equation that makes my life harder

Let x=2X
(4+sqrt(15))^X + (4-sqrt(15))^X = 8^X
An obvious solution is X=1 . So x=2.

3. ## Re: Equation that makes my life harder

Thank you for finding the solution but I have already found it.
I must proof that this equation has only one decision. (I can't proof it using graph because it is enough difficult to construct y=(4+sqrt(15))^X + (4-sqrt(15))^X without, for example, WolframAlpha).

4. ## Re: Equation that makes my life harder

Let $\displaystyle a = 4 + \sqrt{15}$ and $\displaystyle b = 4 - \sqrt{15}$

so that

$\displaystyle a^{x/2} + b^{x/2} = 8^{x/2}$

Multiply this equation by $\displaystyle a^{x/2}$

$\displaystyle a^x + 1 = \left(\sqrt{8a}\right)^x$ since $\displaystyle a b=1$

5. ## Re: Equation that makes my life harder

I'm probably idiot, but I don't really know, how to solve this equation. Sorry.

Let ax=t2

7. ## Re: Equation that makes my life harder

I don't know how to solve this equation either and I don't see how letting $\displaystyle a^x = t^2$ will help.

On the other hand if you graph the function

$\displaystyle f(x) = a^x+1 - \left(\sqrt{8a}\right)^x$ with $\displaystyle x > 0$

you will notice that it is decreasing and that $\displaystyle f(2) = 0$

This shows that $\displaystyle x = 2$ is the only solution

To prove that it is the only solution you would have to show $\displaystyle f ' (x) < 0$

8. ## Re: Equation that makes my life harder

Originally Posted by Idea
I don't know how to solve this equation either and I don't see how letting $\displaystyle a^x = t^2$ will help.

On the other hand if you graph the function

$\displaystyle f(x) = a^x+1 - \left(\sqrt{8a}\right)^x$ with $\displaystyle x > 0$

you will notice that it is decreasing and that $\displaystyle f(2) = 0$

This shows that $\displaystyle x = 2$ is the only solution

To prove that it is the only solution you would have to show $\displaystyle f ' (x) < 0$
My mistake ! You are right.