Finite Math Examples

$8 \cdot 2^{x} = 81 \cdot 3^{x - 1}$

Step 1

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

$\ln (8 \cdot 2^{x}) = \ln (81 \cdot 3^{x - 1})$

Step 2

Expand the left side.

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Step 2.1

Rewrite $\ln (8 \cdot 2^{x})$ as $\ln (8) + \ln (2^{x})$ .

$\ln (8) + \ln (2^{x}) = \ln (81 \cdot 3^{x - 1})$

Step 2.2

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

$\ln (8) + x \ln (2) = \ln (81 \cdot 3^{x - 1})$

Step 3

Expand the right side.

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Step 3.1

Rewrite $\ln (81 \cdot 3^{x - 1})$ as $\ln (81) + \ln (3^{x - 1})$ .

$\ln (8) + x \ln (2) = \ln (81) + \ln (3^{x - 1})$

Step 3.2

Expand $\ln (3^{x - 1})$ by moving $x - 1$ outside the logarithm.

$\ln (8) + x \ln (2) = \ln (81) + (x - 1) \ln (3)$

Step 4

Simplify the right side.

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Step 4.1

Simplify $\ln (81) + (x - 1) \ln (3)$ .

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Step 4.1.1

Simplify each term.

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Step 4.1.1.1

Apply the distributive property.

$\ln (8) + x \ln (2) = \ln (81) + x \ln (3) - 1 \ln (3)$

Step 4.1.1.2

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

$\ln (8) + x \ln (2) = \ln (81) + x \ln (3) - \ln (3)$

Step 4.1.2

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

$\ln (8) + x \ln (2) = x \ln (3) + \ln (\frac{81}{3})$

Step 4.1.3

Divide $81$ by $3$ .

$\ln (8) + x \ln (2) = x \ln (3) + \ln (27)$

Step 5

Reorder $\ln (8)$ and $x \ln (2)$ .

$x \ln (2) + \ln (8) = x \ln (3) + \ln (27)$

Step 6

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

$x \ln (2) + \ln (8) - x \ln (3) - \ln (27) = 0$

Step 7

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

$x \ln (2) - x \ln (3) + \ln (\frac{8}{27}) = 0$

Step 8

Subtract $\ln (\frac{8}{27})$ from both sides of the equation.

$x \ln (2) - x \ln (3) = - \ln (\frac{8}{27})$

Step 9

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

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Step 9.1

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

$x (\ln (2)) - x \ln (3) = - \ln (\frac{8}{27})$

Step 9.2

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

$x (\ln (2)) + x (- 1 \ln (3)) = - \ln (\frac{8}{27})$

Step 9.3

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

$x (\ln (2) - 1 \ln (3)) = - \ln (\frac{8}{27})$

Step 10

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

$x (\ln (2) - \ln (3)) = - \ln (\frac{8}{27})$

Step 11

Divide each term in $x (\ln (2) - \ln (3)) = - \ln (\frac{8}{27})$ by $\ln (2) - \ln (3)$ and simplify.

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Step 11.1

Divide each term in $x (\ln (2) - \ln (3)) = - \ln (\frac{8}{27})$ by $\ln (2) - \ln (3)$ .

$\frac{x (\ln (2) - \ln (3))}{\ln (2) - \ln (3)} = \frac{- \ln (\frac{8}{27})}{\ln (2) - \ln (3)}$

Step 11.2

Simplify the left side.

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Step 11.2.1

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

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Step 11.2.1.1

Cancel the common factor.

$\frac{x (\ln (2) - \ln (3))}{\ln (2) - \ln (3)} = \frac{- \ln (\frac{8}{27})}{\ln (2) - \ln (3)}$

Step 11.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{8}{27})}{\ln (2) - \ln (3)}$

Step 11.3

Simplify the right side.

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Step 11.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{8}{27})}{\ln (2) - \ln (3)}$

Step 12

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{\ln (\frac{8}{27})}{\ln (2) - \ln (3)}$

Decimal Form:

$x = - 3$