# How do you use the graphing calculator to solve 2^(x+6)=2x+15?

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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2
Nov 8, 2017

See below.

#### Explanation:

We will develop an algebraic method instead.

${2}^{x + 6} = 2 x + 15$

This is an equation of type

$a {b}^{x} + c x + d = 0$

that can be solved using the so called Lambert function.

https://en.wikipedia.org/wiki/Lambert_W_function

We will make transformations in order to obtain an equation of type

$\left({c}_{1} y\right) {e}^{{c}_{1} y} = {c}_{2}$ obtaining the solution as

${c}_{1} y = W \left({c}_{2}\right)$ and then

$y = \frac{W \left({c}_{2}\right)}{c} _ 1$

I) Obtaining $\left({c}_{1} y\right) {e}^{{c}_{1} y} = {c}_{2}$ from $a {b}^{x} + c x + d = 0$

making ${b}^{x} = {e}^{x \log b}$ and $y = - \left(c x + d\right)$ we get

$a {e}^{- \left(\frac{y - d}{c}\right) \log b} = y$ or

$a {e}^{\left(\frac{d}{c}\right) \log b} {e}^{- \left(\log \frac{b}{c}\right) y} = y$ or

$a {b}^{\frac{d}{c}} {e}^{- \left(\log \frac{b}{c}\right) y} = y$ or

$a {b}^{\frac{d}{c}} = y {e}^{\left(\log \frac{b}{c}\right) y}$ or

$a {b}^{\frac{d}{c}} \log \frac{b}{c} = \left(\log \frac{b}{c} y\right) {e}^{\left(\log \frac{b}{c}\right) y}$

then here

${c}_{1} = \log \frac{b}{c}$ and ${c}_{2} = a {b}^{\frac{d}{c}} \log \frac{b}{c} = a {b}^{\frac{d}{c}} {c}_{1}$

and then

$y = - \left(c x + d\right) = \frac{W \left(a {b}^{\frac{d}{c}} \log \frac{b}{c}\right)}{\left(\log \frac{b}{c}\right)}$ and finally

$x = - \frac{1}{c} \left(\frac{W \left(a {b}^{\frac{d}{c}} \log \frac{b}{c}\right)}{\left(\log \frac{b}{c}\right)} + d\right)$

Now substituting

$a = {2}^{6} , b = 2 , c = - 2 , d = - 15$

we obtain

$x = - \frac{15 \log 2 + 2 W \left(- \log \frac{2}{4 \sqrt{2}}\right)}{2 \log 2}$ with two solutions

$x = \left\{- 7.2964344147839 , - 2.75297694275627\right\}$

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