Calculus Examples

$8 (4^{5 - x}) = 40$

Step 1

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

$\ln (8 \cdot 4^{5 - x}) = \ln (40)$

Step 2

Expand the left side.

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Step 2.1

Rewrite $\ln (8 \cdot 4^{5 - x})$ as $\ln (8) + \ln (4^{5 - x})$ .

$\ln (8) + \ln (4^{5 - x}) = \ln (40)$

Step 2.2

Expand $\ln (4^{5 - x})$ by moving $5 - x$ outside the logarithm.

$\ln (8) + (5 - x) \ln (4) = \ln (40)$

Step 3

Simplify the left side.

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Step 3.1

Apply the distributive property.

$\ln (8) + 5 \ln (4) - x \ln (4) = \ln (40)$

Step 4

Simplify the left side.

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Step 4.1

Move $5 \ln (4)$ .

$\ln (8) - x \ln (4) + 5 \ln (4) = \ln (40)$

Step 4.2

Reorder $\ln (8)$ and $- x \ln (4)$ .

$- x \ln (4) + \ln (8) + 5 \ln (4) = \ln (40)$

Step 5

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

$- x \ln (4) + \ln (8) + 5 \ln (4) - \ln (40) = 0$

Step 6

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$- x \ln (4) + 5 \ln (4) + \ln (\frac{8}{40}) = 0$

Step 7

Cancel the common factor of $8$ and $40$ .

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Step 7.1

Factor $8$ out of $8$ .

$- x \ln (4) + 5 \ln (4) + \ln (\frac{8 (1)}{40}) = 0$

Step 7.2

Cancel the common factors.

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Step 7.2.1

Factor $8$ out of $40$ .

$- x \ln (4) + 5 \ln (4) + \ln (\frac{8 \cdot 1}{8 \cdot 5}) = 0$

Step 7.2.2

Cancel the common factor.

$- x \ln (4) + 5 \ln (4) + \ln (\frac{8 \cdot 1}{8 \cdot 5}) = 0$

Step 7.2.3

Rewrite the expression.

$- x \ln (4) + 5 \ln (4) + \ln (\frac{1}{5}) = 0$

Step 8

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

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Step 8.1

Subtract $5 \ln (4)$ from both sides of the equation.

$- x \ln (4) + \ln (\frac{1}{5}) = - 5 \ln (4)$

Step 8.2

Subtract $\ln (\frac{1}{5})$ from both sides of the equation.

$- x \ln (4) = - 5 \ln (4) - \ln (\frac{1}{5})$

Step 9

Divide each term in $- x \ln (4) = - 5 \ln (4) - \ln (\frac{1}{5})$ by $- \ln (4)$ and simplify.

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Step 9.1

Divide each term in $- x \ln (4) = - 5 \ln (4) - \ln (\frac{1}{5})$ by $- \ln (4)$ .

$\frac{- x \ln (4)}{- \ln (4)} = \frac{- 5 \ln (4)}{- \ln (4)} + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

Step 9.2

Simplify the left side.

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Step 9.2.1

Dividing two negative values results in a positive value.

$\frac{x \ln (4)}{\ln (4)} = \frac{- 5 \ln (4)}{- \ln (4)} + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

Step 9.2.2

Cancel the common factor of $\ln (4)$ .

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Step 9.2.2.1

Cancel the common factor.

$\frac{x \ln (4)}{\ln (4)} = \frac{- 5 \ln (4)}{- \ln (4)} + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

Step 9.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 5 \ln (4)}{- \ln (4)} + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

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 $\ln (4)$ .

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Step 9.3.1.1.1

Cancel the common factor.

$x = \frac{- 5 \ln (4)}{- \ln (4)} + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

Step 9.3.1.1.2

Rewrite the expression.

$x = \frac{- 5}{- 1} + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

Step 9.3.1.1.3

Move the negative one from the denominator of $\frac{- 5}{- 1}$ .

$x = - 1 \cdot - 5 + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

Step 9.3.1.2

Rewrite $- 1 \cdot - 5$ as $- - 5$ .

$x = - - 5 + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

Step 9.3.1.3

Multiply $- 1$ by $- 5$ .

$x = 5 + \frac{- \ln (\frac{1}{5})}{- \ln (4)}$

Step 9.3.1.4

Dividing two negative values results in a positive value.

$x = 5 + \frac{\ln (\frac{1}{5})}{\ln (4)}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = 5 + \frac{\ln (\frac{1}{5})}{\ln (4)}$

Decimal Form:

$x = 3.83903595 \dots$