Solve for x 3^(2x)-3^(x+2)+8=0

$3^{2 x} - 3^{x + 2} + 8 = 0$

Rewrite $3^{2 x}$ as ${(3^{x})}^{2}$ .

${(3^{x})}^{2} - 3^{x + 2} + 8 = 0$

Rewrite $- 3^{x + 2}$ as $- 3^{2} \cdot 3^{x}$ .

${(3^{x})}^{2} - 3^{2} \cdot 3^{x} + 8 = 0$

Substitute $u$ for $3^{x}$ .

$u^{2} - 3^{2} \cdot u + 8 = 0$

Solve for $u$ .

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Simplify each term.

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Raise $3$ to the power of $2$ .

$u^{2} - 1 \cdot 9 \cdot u + 8 = 0$

Multiply $- 1$ by $9$ .

$u^{2} - 9 u + 8 = 0$

Factor $u^{2} - 9 u + 8$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $8$ and whose sum is $- 9$ .

$- 8, - 1$

Write the factored form using these integers.

$(u - 8) (u - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$u - 8 = 0$

$u - 1 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$u - 8 = 0$

Add $8$ to both sides of the equation.

$u = 8$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

The final solution is all the values that make $(u - 8) (u - 1) = 0$ true.

$u = 8, 1$

Substitute $8$ for $u$ in $u = 3^{x}$ .

$8 = 3^{x}$

Solve $8 = 3^{x}$ .

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Rewrite the equation as $3^{x} = 8$ .

$3^{x} = 8$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (3^{x}) = \ln (8)$

Expand $\ln (3^{x})$ by moving $x$ outside the logarithm.

$x \ln (3) = \ln (8)$

Divide each term by $\ln (3)$ and simplify.

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Divide each term in $x \ln (3) = \ln (8)$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (8)}{\ln (3)}$

Cancel the common factor of $\ln (3)$ .

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Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (8)}{\ln (3)}$

Divide $x$ by $1$ .

$x = \frac{\ln (8)}{\ln (3)}$

Substitute $1$ for $u$ in $u = 3^{x}$ .

$1 = 3^{x}$

Solve $1 = 3^{x}$ .

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Rewrite the equation as $3^{x} = 1$ .

$3^{x} = 1$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (3^{x}) = \ln (1)$

Expand $\ln (3^{x})$ by moving $x$ outside the logarithm.

$x \ln (3) = \ln (1)$

The natural logarithm of $1$ is $0$ .

$x \ln (3) = 0$

Divide each term by $\ln (3)$ and simplify.

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Divide each term in $x \ln (3) = 0$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{0}{\ln (3)}$

Cancel the common factor of $\ln (3)$ .

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Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{0}{\ln (3)}$

Divide $x$ by $1$ .

$x = \frac{0}{\ln (3)}$

Divide $0$ by $\ln (3)$ .

$x = 0$

List the solutions that makes the equation true.

$x = \frac{\ln (8)}{\ln (3)}, 0$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (8)}{\ln (3)}, 0$

Decimal Form:

$x = 1.89278926 \dots, 0$

Solve for x 3^(2x)-3^(x+2)+8=0

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