Finding the real value of x

Question

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 $x$.

Answer 1

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 $x = 8$

Answer 2

Note that $2^{12}=(2^3)^4=8^4$. Equation becomes $$\begin{align} (8^4-8^{-4})(8^4+8^{-4})&=8^x-8^{-x} \\8^8-8^{-8}&=8^x-8^{-x} \\x&=8\qquad\blacksquare\end{align}$$

Answer 3

Let $z=8^x$ and solve

$$z-\frac1z=2^{24}-2^{-24},$$

or

$$z^2-(2^{24}-2^{-24})z-1=0.$$

By inspection (or by the standard formula), the roots are obviously

$$z=2^{24}\text{ and }z=-2^{-24},$$ and there are no others.

The negative root must be rejected ($8^x>0$) and

$$x=\ln_8(2^{24})=\ln_{2^3}(2^{3\cdot8})=8,$$which is guaranteed to be the only real solution.

Answer 4

I think 2^144 = 2^43*3 & 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