solve the equation with superscripts

Question

I need help solving the below equation. First of all, I am not even sure if it can be solved, but I hope it can. $$ 2^{3+x} - 2^{-x} = 2^{3} - 2^{0} $$ Thank you

Answer 1

$8 \cdot 2^x-2^{-x}=7 \\ \\ \text{ multiply } 2^x \text{ on both sides } \\ 8 \cdot 2^{2x}-1=7 \cdot 2^x \\ \text{ you have a quadratic in terms of } 2^x \\ 8 \cdot 2^{2x}-7 \cdot 2^{x}-1=0 \\ (2^x-1)(8 \cdot 2^x+1)=0$

Answer 2

Following Randomgirl's method it is seen that \begin{align} 2^{3+x} = 2^{-x} = 2^{3} - 2^{0} \end{align} first is satisfied if $x=0$, by inspection, and second may be multiplied by $2^{x}$ to obtain \begin{align} 8 \cdot 2^{2x} - 7 \cdot 2^{x} - 1 = 0. \end{align} Let $r= 2^{x}$ to obtain the quadratic equation $8 r^{2} - 7 r -1 =0$ for which the solutions are $r \in \{ 1, - 1/8 \}$. In order to find the value of $x$ it is then required to solve $2^{x} = r$. This is done by using $2^{x} = e^{x \ln(2)}$ for which \begin{align} e^{x \ln(2)} = 1 \hspace{10mm} \rightarrow x = 0 \end{align} and \begin{align} e^{x \ln(2)} = \frac{-1}{8} = e^{\pi i + 2k \pi i - 3 \ln(2)} \hspace{10mm} \rightarrow x_{k} = - 3 + \frac{ (1+2k) \pi i }{ \ln(2) } \hspace{5mm} k \geq 0. \end{align}

Answer 3

another way to do it other than using the quadratic equation is to try to make the inside of each of the parenthases equal 0. For the left side, x must equal 0, which would make it 1-1 or 0. It is actually impossible to make the right one 0, because there is no value of x that could make it -1+1. So, there is only one answer and its 0.