QuestionSolve the equation $2^{2(x+1)} - 9(2^{x}) + 2 = 0$ using the substitution $y = 2^{x}$ .

Studdy Solution

STEP 1

Assumptions1. The given equation is $^{(x+1)}-9\left(^{x}\right)+=0$ . We are using the substitution $y=^{x}$

STEP 2

First, we substitute $y=2^{x}$ into the given equation. This simplifies the equation as $y$ is easier to work with than $2^{x}$ .
$2^{2(x+1)}-9\left(2^{x}\right)+2=0$ becomes $2^{2y}-9y+2=0$

STEP 3

Next, we simplify the equation further by using the property $2^{2y}=(2^{y})^{2}$ , which becomes $y^{2}$ .
So, the equation becomes $y^{2}-9y+2=0$

STEP 4

Now, we solve the quadratic equation $y^{2}-9y+2=0$ by finding its roots. The roots of a quadratic equation $ax^{2}+bx+c=0$ are given by the formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

STEP 5

Substitute $a=1$ , $b=-9$ , and $c=2$ into the formula to find the roots of the equation.
$y=\frac{-(-9)\pm\sqrt{(-9)^{2}-4(1)(2)}}{2(1)}$

STEP 6

implify the equation to find the roots.
$y=\frac{9\pm\sqrt{81-8}}{2}$

STEP 7

Calculate the roots.
$y=\frac{9\pm\sqrt{73}}{2}$

STEP 8

So, the roots of the equation $y^{2}-y+2=0$ are $y=\frac{+\sqrt{73}}{2}$ and $y=\frac{-\sqrt{73}}{2}$ .

STEP 9

Remember that we made the substitution $y=2^{x}$ . So, we substitute $y$ back into the roots to find the solutions for $x$ .
So, the solutions for $x$ are $2^{x}=\frac{9+\sqrt{73}}{2}$ and $2^{x}=\frac{9-\sqrt{73}}{2}$ .

STEP 10

To solve for $x$ , we take the logarithm base2 on both sides of the equation.
For the first solution, we get $x=\log_{2}\left(\frac{9+\sqrt{73}}{2}\right)$ and for the second solution, we get $x=\log_{2}\left(\frac{9-\sqrt{73}}{2}\right)$ These are the solutions for the equation $2^{2(x+)}-9\left(2^{x}\right)+2=0$ .

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