Precalculus Examples

$2^{2 x + 10} = 3^{x - 40}$

Step 1

Take the log of both sides of the equation.

$\ln (2^{2 x + 10}) = \ln (3^{x - 40})$

Step 2

Expand $\ln (2^{2 x + 10})$ by moving $2 x + 10$ outside the logarithm.

$(2 x + 10) \ln (2) = \ln (3^{x - 40})$

Step 3

Expand $\ln (3^{x - 40})$ by moving $x - 40$ outside the logarithm.

$(2 x + 10) \ln (2) = (x - 40) \ln (3)$

Step 4

Solve the equation for $x$ .

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Step 4.1

Simplify $(2 x + 10) \ln (2)$ .

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Step 4.1.1

Rewrite.

$0 + 0 + (2 x + 10) \ln (2) = (x - 40) \ln (3)$

Step 4.1.2

Simplify by adding zeros.

$(2 x + 10) \ln (2) = (x - 40) \ln (3)$

Step 4.1.3

Apply the distributive property.

$2 x \ln (2) + 10 \ln (2) = (x - 40) \ln (3)$

Step 4.2

Apply the distributive property.

$2 x \ln (2) + 10 \ln (2) = x \ln (3) - 40 \ln (3)$

Step 4.3

Subtract $x \ln (3)$ from both sides of the equation.

$2 x \ln (2) + 10 \ln (2) - x \ln (3) = - 40 \ln (3)$

Step 4.4

Subtract $10 \ln (2)$ from both sides of the equation.

$2 x \ln (2) - x \ln (3) = - 40 \ln (3) - 10 \ln (2)$

Step 4.5

Factor $x$ out of $2 x \ln (2) - x \ln (3)$ .

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Step 4.5.1

Factor $x$ out of $2 x \ln (2)$ .

$x (2 \ln (2)) - x \ln (3) = - 40 \ln (3) - 10 \ln (2)$

Step 4.5.2

Factor $x$ out of $- x \ln (3)$ .

$x (2 \ln (2)) + x (- 1 \ln (3)) = - 40 \ln (3) - 10 \ln (2)$

Step 4.5.3

Factor $x$ out of $x (2 \ln (2)) + x (- 1 \ln (3))$ .

$x (2 \ln (2) - 1 \ln (3)) = - 40 \ln (3) - 10 \ln (2)$

Step 4.6

Rewrite $- 1 \ln (3)$ as $- \ln (3)$ .

$x (2 \ln (2) - \ln (3)) = - 40 \ln (3) - 10 \ln (2)$

Step 4.7

Divide each term in $x (2 \ln (2) - \ln (3)) = - 40 \ln (3) - 10 \ln (2)$ by $2 \ln (2) - \ln (3)$ and simplify.

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Step 4.7.1

Divide each term in $x (2 \ln (2) - \ln (3)) = - 40 \ln (3) - 10 \ln (2)$ by $2 \ln (2) - \ln (3)$ .

$\frac{x (2 \ln (2) - \ln (3))}{2 \ln (2) - \ln (3)} = \frac{- 40 \ln (3)}{2 \ln (2) - \ln (3)} + \frac{- 10 \ln (2)}{2 \ln (2) - \ln (3)}$

Step 4.7.2

Simplify the left side.

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Step 4.7.2.1

Cancel the common factor of $2 \ln (2) - \ln (3)$ .

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Step 4.7.2.1.1

Cancel the common factor.

$\frac{x (2 \ln (2) - \ln (3))}{2 \ln (2) - \ln (3)} = \frac{- 40 \ln (3)}{2 \ln (2) - \ln (3)} + \frac{- 10 \ln (2)}{2 \ln (2) - \ln (3)}$

Step 4.7.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 40 \ln (3)}{2 \ln (2) - \ln (3)} + \frac{- 10 \ln (2)}{2 \ln (2) - \ln (3)}$

Step 4.7.3

Simplify the right side.

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Step 4.7.3.1

Combine the numerators over the common denominator.

$x = \frac{- 40 \ln (3) - 10 \ln (2)}{2 \ln (2) - \ln (3)}$

Step 4.7.3.2

Factor $10$ out of $- 40 \ln (3) - 10 \ln (2)$ .

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Step 4.7.3.2.1

Factor $10$ out of $- 40 \ln (3)$ .

$x = \frac{10 (- 4 \ln (3)) - 10 \ln (2)}{2 \ln (2) - \ln (3)}$

Step 4.7.3.2.2

Factor $10$ out of $- 10 \ln (2)$ .

$x = \frac{10 (- 4 \ln (3)) + 10 (- \ln (2))}{2 \ln (2) - \ln (3)}$

Step 4.7.3.2.3

Factor $10$ out of $10 (- 4 \ln (3)) + 10 (- \ln (2))$ .

$x = \frac{10 (- 4 \ln (3) - \ln (2))}{2 \ln (2) - \ln (3)}$

Step 4.7.3.3

Factor $- 1$ out of $- 4 \ln (3)$ .

$x = \frac{10 (- (4 \ln (3)) - \ln (2))}{2 \ln (2) - \ln (3)}$

Step 4.7.3.4

Factor $- 1$ out of $- \ln (2)$ .

$x = \frac{10 (- (4 \ln (3)) - \ln (2))}{2 \ln (2) - \ln (3)}$

Step 4.7.3.5

Factor $- 1$ out of $- (4 \ln (3)) - \ln (2)$ .

$x = \frac{10 (- (4 \ln (3) + \ln (2)))}{2 \ln (2) - \ln (3)}$

Step 4.7.3.6

Simplify the expression.

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Step 4.7.3.6.1

Rewrite $- (4 \ln (3) + \ln (2))$ as $- 1 (4 \ln (3) + \ln (2))$ .

$x = \frac{10 (- 1 (4 \ln (3) + \ln (2)))}{2 \ln (2) - \ln (3)}$

Step 4.7.3.6.2

Move the negative in front of the fraction.

$x = - \frac{10 (4 \ln (3) + \ln (2))}{2 \ln (2) - \ln (3)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{10 (4 \ln (3) + \ln (2))}{2 \ln (2) - \ln (3)}$

Decimal Form:

$x = - 176.84787556 \dots$