Precalculus Examples

$8 (4^{6 - 2 x}) + 13 = 41$

Step 1

Multiply $8 \cdot 4^{6 - 2 x}$ .

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Step 1.1

Rewrite $8$ as $2^{3}$ .

$2^{3} \cdot 4^{6 - 2 x} + 13 = 41$

Step 1.2

Rewrite $4$ as $2^{2}$ .

$2^{3} \cdot {(2^{2})}^{6 - 2 x} + 13 = 41$

Step 1.3

Multiply the exponents in ${(2^{2})}^{6 - 2 x}$ .

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Step 1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$2^{3} \cdot 2^{2 (6 - 2 x)} + 13 = 41$

Step 1.3.2

Apply the distributive property.

$2^{3} \cdot 2^{2 \cdot 6 + 2 (- 2 x)} + 13 = 41$

Step 1.3.3

Multiply $2$ by $6$ .

$2^{3} \cdot 2^{12 + 2 (- 2 x)} + 13 = 41$

Step 1.3.4

Multiply $- 2$ by $2$ .

$2^{3} \cdot 2^{12 - 4 x} + 13 = 41$

Step 1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2^{3 + 12 - 4 x} + 13 = 41$

Step 1.5

Add $3$ and $12$ .

$2^{15 - 4 x} + 13 = 41$

Step 2

Move all terms not containing $x$ to the right side of the equation.

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Step 2.1

Subtract $13$ from both sides of the equation.

$2^{15 - 4 x} = 41 - 13$

Step 2.2

Subtract $13$ from $41$ .

$2^{15 - 4 x} = 28$

Step 3

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

$\ln (2^{15 - 4 x}) = \ln (28)$

Step 4

Expand $\ln (2^{15 - 4 x})$ by moving $15 - 4 x$ outside the logarithm.

$(15 - 4 x) \ln (2) = \ln (28)$

Step 5

Simplify the left side.

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Step 5.1

Apply the distributive property.

$15 \ln (2) - 4 x \ln (2) = \ln (28)$

Step 6

Reorder $15 \ln (2)$ and $- 4 x \ln (2)$ .

$- 4 x \ln (2) + 15 \ln (2) = \ln (28)$

Step 7

Move all the terms containing a logarithm to the left side of the equation.

$- 4 x \ln (2) + 15 \ln (2) - \ln (28) = 0$

Step 8

Move all terms not containing $x$ to the right side of the equation.

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Step 8.1

Subtract $15 \ln (2)$ from both sides of the equation.

$- 4 x \ln (2) - \ln (28) = - 15 \ln (2)$

Step 8.2

Add $\ln (28)$ to both sides of the equation.

$- 4 x \ln (2) = - 15 \ln (2) + \ln (28)$

Step 9

Divide each term in $- 4 x \ln (2) = - 15 \ln (2) + \ln (28)$ by $- 4 \ln (2)$ and simplify.

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Step 9.1

Divide each term in $- 4 x \ln (2) = - 15 \ln (2) + \ln (28)$ by $- 4 \ln (2)$ .

$\frac{- 4 x \ln (2)}{- 4 \ln (2)} = \frac{- 15 \ln (2)}{- 4 \ln (2)} + \frac{\ln (28)}{- 4 \ln (2)}$

Step 9.2

Simplify the left side.

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Step 9.2.1

Cancel the common factor of $- 4$ .

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Step 9.2.1.1

Cancel the common factor.

$\frac{- 4 x \ln (2)}{- 4 \ln (2)} = \frac{- 15 \ln (2)}{- 4 \ln (2)} + \frac{\ln (28)}{- 4 \ln (2)}$

Step 9.2.1.2

Rewrite the expression.

$\frac{x \ln (2)}{\ln (2)} = \frac{- 15 \ln (2)}{- 4 \ln (2)} + \frac{\ln (28)}{- 4 \ln (2)}$

Step 9.2.2

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

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Step 9.2.2.1

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{- 15 \ln (2)}{- 4 \ln (2)} + \frac{\ln (28)}{- 4 \ln (2)}$

Step 9.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 15 \ln (2)}{- 4 \ln (2)} + \frac{\ln (28)}{- 4 \ln (2)}$

Step 9.3

Simplify the right side.

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Step 9.3.1

Simplify each term.

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Step 9.3.1.1

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

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Step 9.3.1.1.1

Cancel the common factor.

$x = \frac{- 15 \ln (2)}{- 4 \ln (2)} + \frac{\ln (28)}{- 4 \ln (2)}$

Step 9.3.1.1.2

Rewrite the expression.

$x = \frac{- 15}{- 4} + \frac{\ln (28)}{- 4 \ln (2)}$

Step 9.3.1.2

Dividing two negative values results in a positive value.

$x = \frac{15}{4} + \frac{\ln (28)}{- 4 \ln (2)}$

Step 9.3.1.3

Move the negative in front of the fraction.

$x = \frac{15}{4} - \frac{\ln (28)}{4 \ln (2)}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = \frac{15}{4} - \frac{\ln (28)}{4 \ln (2)}$

Decimal Form:

$x = 2.54816126 \dots$