# Can you help me find x? (Exponential Function).

## ${2}^{x} - 10 \left({2}^{-} x\right) + 3 = 0$

Apr 13, 2017

$x = 1$

#### Explanation:

Multiplying each term of ${2}^{x} - 10 \left({2}^{-} x\right) + 3 = 0$ by ${2}^{x}$, we gwt

${2}^{2 x} - 10 + 3 \times {2}^{x} = 0$.

Now let ${2}^{x} = u$, the above becomes

${u}^{2} + 3 u - 10 = 0$

or ${u}^{2} + 5 u - 2 u - 10 = 0$

or $u \left(u + 5\right) - 2 \left(u + 5\right) = 0$

or $\left(u - 2\right) \left(u + 5\right) = 0$

but as $u = {2}^{x}$, $u + 5 \ne 0$ and dividing both sides by it, we get

$u = 2$ or ${2}^{x} = 2$ or $x = 1$

Apr 13, 2017

$x = 1$

#### Explanation:

Write as:$\text{ "2^x-10/2^x+3" "=" } 0$

$\text{ "color(green)([2^xcolor(red)(xx1)] -10/2^x+[3color(red)(xx1)]" "=" } 0$

" "color(green)([2^xcolor(red)(xx2^x/2^x)] -10/2^x+[3color(red)(xx2^x/2^x)]=0

$\text{ "(2^(2x))/2^x" " -10/2^x+" "(3xx2^x)/2^x" "=" } 0$

$\text{ } {2}^{2 x} + \left(3 \times {2}^{x}\right) - 10 = 0$

Changed my mind about the format. Write ${2}^{2 x} \text{ as } {\left({2}^{x}\right)}^{2}$

Set $a = {2}^{x}$ giving

${a}^{2} + 3 a - 10 = 0$

$\left(a - 2\right) \left(a + 5\right) = 0$

$a = - 2 \mathmr{and} a = + 5$

$\implies {2}^{x} = + 2 \mathmr{and} {2}^{x} = - 5$

Using logs

Consider ${2}^{x} = 2$

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

Consider ${2}^{x} = - 5$

$x \ln \left(2\right) = \ln \left(- 5\right)$

but $\ln \left(- 5\right)$ is undefined so not a solution.

Thus $x = 1$