Precalculus Examples

$49^{x} - 6 \cdot 7^{x} = - 9$

Step 1

Add $9$ to both sides of the equation.

$49^{x} - 6 \cdot 7^{x} + 9 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Rewrite $49^{x}$ as ${(7^{2})}^{x}$ .

${(7^{2})}^{x} - 6 \cdot 7^{x} + 9 = 0$

Step 2.2

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

${(7^{x})}^{2} - 6 \cdot 7^{x} + 9 = 0$

Step 2.3

Let $u = 7^{x}$ . Substitute $u$ for all occurrences of $7^{x}$ .

$u^{2} - 6 u + 9 = 0$

Step 2.4

Factor using the perfect square rule.

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Step 2.4.1

Rewrite $9$ as $3^{2}$ .

$u^{2} - 6 u + 3^{2} = 0$

Step 2.4.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Step 2.4.3

Rewrite the polynomial.

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

Step 2.4.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 3$ .

${(u - 3)}^{2} = 0$

Step 2.5

Replace all occurrences of $u$ with $7^{x}$ .

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

Step 3

Set the $7^{x} - 3$ equal to $0$ .

$7^{x} - 3 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $3$ to both sides of the equation.

$7^{x} = 3$

Step 4.2

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

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

Step 4.3

Expand $\ln (7^{x})$ by moving $x$ outside the logarithm.

$x \ln (7) = \ln (3)$

Step 4.4

Divide each term in $x \ln (7) = \ln (3)$ by $\ln (7)$ and simplify.

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Step 4.4.1

Divide each term in $x \ln (7) = \ln (3)$ by $\ln (7)$ .

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (3)}{\ln (7)}$

Step 4.4.2

Simplify the left side.

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Step 4.4.2.1

Cancel the common factor of $\ln (7)$ .

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Step 4.4.2.1.1

Cancel the common factor.

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (3)}{\ln (7)}$

Step 4.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (3)}{\ln (7)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (3)}{\ln (7)}$

Decimal Form:

$x = 0.56457503 \dots$