Question

Exponential Equation WITHOUT CALCULATOR

Answer 1

Multiply both sides of the first equation by $2^x$:

$2^x-2^{-x}=4\implies 2^{2x}-1=4\cdot2^x\implies 2^{2x}-4\cdot2^x-1=0$

This is quadratic in $2^x$; to make this clear, substitute $y=2^x$. Then

$y^2-4y-1=0\implies y=2\pm\sqrt5$

One of these solutions for $y$ is negative. But, if $x$ is real, then $y=2^x$ is always supposed to be positive, so we can throw out the negative root, leaving $y=2+\sqrt5$.

We actually don't have to solve for $x$ exactly. We can just rewrite the next two equations in terms of $y$.


$2^{2x}+2^{-2x}=(2^x)^2+(2^{-x})^2=y^2+\dfrac1{y^2}$

$2^{3x}-2^{-3x}=y^3-\dfrac1{y^3}$


Since $y=2+\sqrt5$, we get

$(2+\sqrt5)^2+\dfrac1{(2+\sqrt5)^2}=18$

$(2+\sqrt5)^3-\dfrac1{(2+\sqrt5)^3}=76$