# How do you solve 2^x= (2^-x)-(3/2)?

Nov 24, 2015

$x = - 1$

#### Explanation:

We will be using the following:

$a {x}^{2} + b x + c = 0 \implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\ln \left({a}^{b}\right) = b \ln \left(a\right)$

${2}^{x} = {2}^{- x} - \frac{3}{2}$

Multiplying both sides by ${2}^{x}$ gives

${\left({2}^{x}\right)}^{2} = 1 - \left(\frac{3}{2}\right) {2}^{x}$

$\implies {\left({2}^{x}\right)}^{2} + \left(\frac{3}{2}\right) {2}^{x} - 1 = 0$

This is now a quadratic equation. Applying the quadratic formula, we obtain

${2}^{x} = \frac{- \frac{3}{2} \pm \sqrt{\frac{9}{4} + 4}}{2} = \frac{- \frac{3}{2} \pm \frac{5}{2}}{2} = \frac{- 3 \pm 5}{4}$

We know, however, that for all real-valued $x$, ${2}^{x} > 0$ and so we can discard ${2}^{x} = - 2$ leaving us with ${2}^{x} = \frac{1}{2}$

Now, taking the natural log of both sides and applying the property of logarithms stated above,

$\ln \left({2}^{x}\right) = \ln \left(\frac{1}{2}\right)$

$\implies x \ln \left(2\right) = \ln \left(\frac{1}{2}\right)$

$\implies x = \ln \frac{\frac{1}{2}}{\ln} \left(2\right)$

For more complicated values, we would be done here, however as $\frac{1}{2} = {2}^{- 1}$

$x = \ln \frac{{2}^{- 1}}{\ln \left(2\right)} = \frac{- 1 \ln \left(2\right)}{\ln \left(2\right)} = - 1$

Thus $x = - 1$

(Note that observing earlier that $\frac{1}{2} = {2}^{-} 1$ would have allowed us to say that $x = - 1$ without any work with logarithms, however the above is a more generally applicable technique which works even when the values do not work out so nicely)