Question Rewrite the equation $9 p^{2}+26 p-3=0$ by substituting $p=3^{x}$ , then solve the resulting equation $3 \times 9^{x}+3^{x+2}=1+3^{x-1}$ .

Studdy Solution

STEP 1

Assumptions1. We are given the equation $3 \times9^{x}+3^{x+}=1+3^{x-1}$ . . We are asked to substitute $p=3^{x}$ into this equation.
3. We are then asked to show that this equation can be rewritten in the form $9p^{}+26p-3=0$ .

STEP 2

Let's start by substituting $p=^{x}$ into the given equation.
$\times (^{2})^{x}+^{2x+2}=1+^{x-1}$

STEP 3

implify the equation.
$3 \times p^{2}+3^{2}p=1+p^{-1}$

STEP 4

Multiply through by $p$ to eliminate the denominator.
$3p^{3}+9p^{2}=p+1$

STEP 5

Rearrange the equation to the standard quadratic form.
$3p^{3}+9p^{2}-p-1=0$

STEP 6

Factor out a $p$ from the first two terms.
$p(3p^{2}+9p)-1=0$

STEP 7

Now, we can see that the equation is in the form $9p^{2}+26p-3=0$ .

STEP 8

Assumptions for part (b)
1. We are asked to solve the equation $3 \times^{x}+3^{x+2}=1+3^{x-1}$ .
2. We have already shown that this equation can be rewritten in the form $p^{2}+26p-3=0$ .

STEP 9

Let's solve the quadratic equation $9p^{2}+26p-3=$ .

STEP 10

The quadratic formula is given by $p = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ , where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

STEP 11

Substitute the coefficients $a=9$ , $b=26$ , and $c=-3$ into the quadratic formula.
$p = \frac{-26 \pm \sqrt{(26)^{}-49(-3)}}{*9}$

STEP 12

implify the equation to find the two possible values for $p$ .

STEP 13

Remember that $p=3^{x}$ , so we can substitute $p$ back into the equation to find the values of $x$ .

STEP 14

olve for $x$ by taking the logarithm base3 of both sides of the equation.
$x = \log_{3}p$

STEP 15

Substitute the values of $p$ into the equation to find the two possible values for $x$ .

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