# Thread: I have been stuck on this for a bit now

1. ## I have been stuck on this for a bit now

Find all values x such that $\displaystyle (2^x-4)^3+(4^x-2)^3 = (4^x+2^x-6)^3$

I expanded the entire thing:

$\displaystyle 2^3^x+2^2^x-38+4^3^x+4^2^x+4^x = (4^3^x)(2^5^x)+(4^2^x)(2^3^x)+4^x+4^x-105+2^x+2^x$

but not quite sure what to do from here. Is there a logarithm I can use? Or would I have to go further and get it all to one side equaling to 0?

2. ## Re: I have been stuck on this for a bit now

Originally Posted by brokaliv
Find all values x such that $\displaystyle (2^x-4)^3+(4^x-2)^3 = (4^x+2^x-6)^3$

I expanded the entire thing:

$\displaystyle 2^3^x+2^2^x-38+4^3^x+4^2^x+4^x = (4^3^x)(2^5^x)+(4^2^x)(2^3^x)+4^x+4^x-105+2^x+2^x$

but not quite sure what to do from here. Is there a logarithm I can use? Or would I have to go further and get it all to one side equaling to 0?
I don't think you can have expanded these out correctly... Remember \displaystyle \displaystyle \begin{align*} (a + b)^3 = a^3 + 3a^2b + 3a\,b^2 + b^3 \end{align*}...

3. ## Re: I have been stuck on this for a bit now

Let a=2^x-4 and b=4^x-2 so we are solving a^3+b^3=(a+b)^3
Multiply out the bracket and get 3ab(a+b)=0 So a=0 or b=0 or a+b=0 giving x= 1/2 x=1 and x=2

4. ## Re: I have been stuck on this for a bit now

>_< These unknown exponents are really screwing with my head, I can't seem to get it expanded correctly.

I'm going to attempt biffboy's method to see if I can get it to work that way.

Btw, I graphed both sides out and they it the x-axis at 1; but it doesn't show the other values that way either. :/

Thank you both for the help and advice so far. Much appreciated.

5. ## Re: I have been stuck on this for a bit now

Originally Posted by biffboy
Let a=2^x-4 and b=4^x-2 so we are solving a^3+b^3=(a+b)^3
Multiply out the bracket and get 3ab(a+b)=0 So a=0 or b=0 or a+b=0 giving x= 1/2 x=1 and x=2
$\displaystyle a=2^x-4$
$\displaystyle b=4^x-2$

therefore,

$\displaystyle a^3+b^3=(a+b)^3$

I then expanded the right side

$\displaystyle a^3+b^3=a^3+3a^2b+3ab^2+b^3$

So I have to set it to 0 so I have to subtract $\displaystyle a^3$ and $\displaystyle b^3$ getting:

$\displaystyle 3a^2b+3ab^2=0$

and then I factor out $\displaystyle 3ab$

$\displaystyle 3ab(a+b)=0$

I did the following and got your same answers, but don't I have to use logarithms?

$\displaystyle 0=2^x-4$
$\displaystyle 2^x = 4$
$\displaystyle x=2$

$\displaystyle 0=4^x-2$
$\displaystyle 4^x = 2$
$\displaystyle x = 1/2$

problem I am having is if $\displaystyle a+b=0$

I assume I resubstitute the originals back in?

$\displaystyle 2^x-4+4^x-2=0$
$\displaystyle 2^x+4^x-6=0$
$\displaystyle 2^x+4^x=6$
$\displaystyle 6^x=6$
$\displaystyle x=1$

Should I be using Logarithm's to solve these when they get to the form $\displaystyle 6^x=6$ or am I to just figure it out since its a simple guess?

6. ## Re: I have been stuck on this for a bit now

This is correct now. You will get the right answers if you use logs but it is not necessary if you can see the answer without using them, as in 2^x=4 and 6^x=6.
For 4^x=2 we see that 2 is the square root of 4 so x=1/2. Alternatively log(4^x)=log2 so xlog4=log2 so x= log2/log4= 0.5

7. ## Re: I have been stuck on this for a bit now

Could have sworn I said thank you both for the help...apparently cell phone didn't post it O_o

Thank you both!