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

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

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Answer 1 · 2017-11-08T22:10:20

See below.

Explanation:

We will develop an algebraic method instead.

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

This is an equation of type

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

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

$(c_{1} y) e_{1}^{c_{1} y} = c_{2}$ $(c_1 y)e^(c_1 y) = c_2$ obtaining the solution as

$c_{1} y = W (c_{2})$ $c_1 y = W(c_2)$ and then

$y = \frac{W (c_{2})}{c_{1}}$ $y = (W(c_2))/c_1$

I) Obtaining $(c_{1} y) e_{1}^{c_{1} y} = c_{2}$ $(c_1 y)e^(c_1 y) = c_2$ from $a b^{x} + c x + d = 0$ $a b^x+c x+d=0$

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

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

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

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

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

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

then here

$c_{1} = \frac{log b}{c}$ $c_1 = logb/c$ and $c_{2} = a b^{\frac{d}{c}} \frac{log b}{c} = a b^{\frac{d}{c}} c_{1}$ $c_2 = a b^(d/c) logb/c = a b^(d/c) c_1$

and then

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

$x = - \frac{1}{c} ⎛ ⎜ ⎝ \frac{W (a b^{\frac{d}{c}} \frac{log b}{c})}{(\frac{log b}{c})} + d ⎞ ⎟ ⎠$ $x = -1/c( (W(a b^(d/c) logb/c ))/(( logb/c))+d)$

Now substituting

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

we obtain

$x = - \frac{15 log 2 + 2 W (- \frac{log 2}{4 \sqrt{2}})}{2 log 2}$ $x =- (15 log2 +2 W(-log2/(4 sqrt[2])))/(2 log2)$ with two solutions

$x = {- 7.2964344147839, - 2.75297694275627}$ $x = {-7.2964344147839,-2.75297694275627}$

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

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