Precalculus Examples

7^{2 x} + 7^{x} - 6 = 0

$7^{2 x} + 7^{x} - 6 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite

7^{2 x}

$7^{2 x}$ as

{(7^{x})}^{2}

${(7^{x})}^{2}$ .

{(7^{x})}^{2} + 7^{x} - 6 = 0

${(7^{x})}^{2} + 7^{x} - 6 = 0$

Step 1.2

Let

u = 7^{x}

$u = 7^{x}$ . Substitute

u

$u$ for all occurrences of

7^{x}

$7^{x}$ .

u^{2} + u - 6 = 0

$u^{2} + u - 6 = 0$

Step 1.3

Factor

u^{2} + u - 6

$u^{2} + u - 6$ using the AC method.

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Step 1.3.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 6

$- 6$ and whose sum is

1

$1$ .

- 2, 3

$- 2, 3$

Step 1.3.2

Write the factored form using these integers.

(u - 2) (u + 3) = 0

$(u - 2) (u + 3) = 0$

(u - 2) (u + 3) = 0

$(u - 2) (u + 3) = 0$

Step 1.4

Replace all occurrences of

u

$u$ with

7^{x}

$7^{x}$ .

(7^{x} - 2) (7^{x} + 3) = 0

$(7^{x} - 2) (7^{x} + 3) = 0$

(7^{x} - 2) (7^{x} + 3) = 0

$(7^{x} - 2) (7^{x} + 3) = 0$

Step 2

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

7^{x} - 2 = 0

$7^{x} - 2 = 0$

7^{x} + 3 = 0

$7^{x} + 3 = 0$

Step 3

Set

7^{x} - 2

$7^{x} - 2$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.1

Set

7^{x} - 2

$7^{x} - 2$ equal to

0

$0$ .

7^{x} - 2 = 0

$7^{x} - 2 = 0$

Step 3.2

Solve

7^{x} - 2 = 0

$7^{x} - 2 = 0$ for

x

$x$ .

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Step 3.2.1

Add

2

$2$ to both sides of the equation.

7^{x} = 2

$7^{x} = 2$

Step 3.2.2

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

\ln (7^{x}) = \ln (2)

$\ln (7^{x}) = \ln (2)$

Step 3.2.3

Expand

\ln (7^{x})

$\ln (7^{x})$ by moving

x

$x$ outside the logarithm.

x \ln (7) = \ln (2)

$x \ln (7) = \ln (2)$

Step 3.2.4

Divide each term in

x \ln (7) = \ln (2)

$x \ln (7) = \ln (2)$ by

\ln (7)

$\ln (7)$ and simplify.

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Step 3.2.4.1

Divide each term in

x \ln (7) = \ln (2)

$x \ln (7) = \ln (2)$ by

\ln (7)

$\ln (7)$ .

\frac{x \ln (7)}{\ln (7)} = \frac{\ln (2)}{\ln (7)}

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (2)}{\ln (7)}$

Step 3.2.4.2

Simplify the left side.

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Step 3.2.4.2.1

Cancel the common factor of

\ln (7)

$\ln (7)$ .

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Step 3.2.4.2.1.1

Cancel the common factor.

\frac{x \ln (7)}{\ln (7)} = \frac{\ln (2)}{\ln (7)}

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (2)}{\ln (7)}$

Step 3.2.4.2.1.2

Divide

x

$x$ by

1

$1$ .

x = \frac{\ln (2)}{\ln (7)}

$x = \frac{\ln (2)}{\ln (7)}$

x = \frac{\ln (2)}{\ln (7)}

$x = \frac{\ln (2)}{\ln (7)}$

x = \frac{\ln (2)}{\ln (7)}

$x = \frac{\ln (2)}{\ln (7)}$

x = \frac{\ln (2)}{\ln (7)}

$x = \frac{\ln (2)}{\ln (7)}$

x = \frac{\ln (2)}{\ln (7)}

$x = \frac{\ln (2)}{\ln (7)}$

x = \frac{\ln (2)}{\ln (7)}

$x = \frac{\ln (2)}{\ln (7)}$

Step 4

Set

7^{x} + 3

$7^{x} + 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

7^{x} + 3

$7^{x} + 3$ equal to

0

$0$ .

7^{x} + 3 = 0

$7^{x} + 3 = 0$

Step 4.2

Solve

7^{x} + 3 = 0

$7^{x} + 3 = 0$ for

x

$x$ .

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Step 4.2.1

Subtract

3

$3$ from both sides of the equation.

7^{x} = - 3

$7^{x} = - 3$

Step 4.2.2

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

\ln (7^{x}) = \ln (- 3)

$\ln (7^{x}) = \ln (- 3)$

Step 4.2.3

The equation cannot be solved because

\ln (- 3)

$\ln (- 3)$ is undefined.

Undefined

Step 4.2.4

There is no solution for

7^{x} = - 3

$7^{x} = - 3$

No solution

Step 5

The final solution is all the values that make

(7^{x} - 2) (7^{x} + 3) = 0

$(7^{x} - 2) (7^{x} + 3) = 0$ true.

x = \frac{\ln (2)}{\ln (7)}

$x = \frac{\ln (2)}{\ln (7)}$

Step 6

The result can be shown in multiple forms.

Exact Form:

x = \frac{\ln (2)}{\ln (7)}

$x = \frac{\ln (2)}{\ln (7)}$

Decimal Form:

x = 0.35620718 \dots

$x = 0.35620718 \dots$