Finding the real value of x I replaced the numbers with variables so i let $y=2$ and $z=12$ and i got the following after simplifying the left side:

$$2^{144}-\frac{1}{2^{144}}=8^x-8^{-x}$$

but i'm not seeing how this can help me find $x$.
The answer is $x = 8$

$$(2^{12}-2^{-12})(2^{12}+2^{-12})=8^x-8^{-x}$$

=$$(2^{12})^2-(2^{-12})^2=8^x-8^{-x}  $$

=$$\frac{2^{48}-1}{2^{24}}=\frac{8^{2x}-1}{8^x}$$

=$$\frac{2^{48}-1}{2^{24}}=\frac{2^{3(2x)}-1}{2^{3(x)}}$$

Now since the equation is monotonic we know there is a one-to-one correspondence we can write 

$${48}=3(2x) \ \ or \ \ {24}= {3(x)} $$

Either case $x = 8$
I think 
2^144 = 2^43*3 
&amp; 8^x = 2^3^x = 2^3x

so 
2^43*3 - 1/ 2^43*3  = 2^3x - 1/ 2^3x

by comparing you will get x = 43