(2^x) - 6/(2^-x)=6
$\displaystyle \displaystyle \begin{align*} 2^x - \frac{6}{2^{-x}} &= 6 \\ 2^x - 6\cdot 2^x &= 6 \\ -5\cdot 2^x &= 6 \\ 2^x &= -\frac{6}{5} \end{align*} $
This does not have any solutions in the real numbers, because $\displaystyle \displaystyle \begin{align*} 2^x > 0 \end{align*} $ for all $\displaystyle \displaystyle \begin{align*} x \end{align*} $.
Hello, sluggerbroth!
(2^x) - 6/(2^-x) = 6
I suspect that the problem is: .$\displaystyle 2^x - \frac{6}{2^x} \:=\:6$
Multiply by $\displaystyle 2^x:\;(2^x)^2 - 6 \:=\:6\cdot2^x$
And we have: .$\displaystyle (2^x)^2 - 6\cdot2^x - 6 \:=\:0$ . . . a quadratic in $\displaystyle 2^x$
Quadratic Formula: .$\displaystyle 2^x \:=\:\frac{6\pm\sqrt{60}}{2} \:=\:3\pm\sqrt{15}$
Take logs, base 10: .$\displaystyle \log(2^x) \:=\:\log(3\pm\sqrt{15}) $
. . . . . . . . . . . . . . $\displaystyle x\cdot\log(2) \:=\:\log(3\pm\sqrt{15})$
n . . . . . . . . . . . . . . . . . . $\displaystyle \boxed{\;x \:=\:\frac{\log(3 + \sqrt{15})}{\log(2)}\;} $
Note that $\displaystyle 3 - \sqrt{15}$ is negative.