Algebra Examples

$4^{5 x + 2} = 0.8 \cdot 5^{5 x + 2}$

Step 1

Take the log of both sides of the equation.

$\ln (4^{5 x + 2}) = \ln (0.8 \cdot 5^{5 x + 2})$

Step 2

Expand $\ln (4^{5 x + 2})$ by moving $5 x + 2$ outside the logarithm.

$(5 x + 2) \ln (4) = \ln (0.8 \cdot 5^{5 x + 2})$

Step 3

Rewrite $\ln (0.8 \cdot 5^{5 x + 2})$ as $\ln (0.8) + \ln (5^{5 x + 2})$ .

$(5 x + 2) \ln (4) = \ln (0.8) + \ln (5^{5 x + 2})$

Step 4

Expand $\ln (5^{5 x + 2})$ by moving $5 x + 2$ outside the logarithm.

$(5 x + 2) \ln (4) = \ln (0.8) + (5 x + 2) \ln (5)$

Step 5

Solve the equation for $x$ .

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Step 5.1

Simplify the left side.

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Step 5.1.1

Simplify $(5 x + 2) \ln (4)$ .

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Step 5.1.1.1

Apply the distributive property.

$5 x \ln (4) + 2 \ln (4) = \ln (0.8) + (5 x + 2) \ln (5)$

Step 5.1.1.2

Simplify $5 \ln (4)$ by moving $5$ inside the logarithm.

$x \ln (4^{5}) + 2 \ln (4) = \ln (0.8) + (5 x + 2) \ln (5)$

Step 5.1.1.3

Simplify $2 \ln (4)$ by moving $2$ inside the logarithm.

$x \ln (4^{5}) + \ln (4^{2}) = \ln (0.8) + (5 x + 2) \ln (5)$

Step 5.1.1.4

Simplify each term.

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Step 5.1.1.4.1

Raise $4$ to the power of $5$ .

$x \ln (1024) + \ln (4^{2}) = \ln (0.8) + (5 x + 2) \ln (5)$

Step 5.1.1.4.2

Raise $4$ to the power of $2$ .

$x \ln (1024) + \ln (16) = \ln (0.8) + (5 x + 2) \ln (5)$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $\ln (0.8) + (5 x + 2) \ln (5)$ .

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Step 5.2.1.1

Simplify each term.

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Step 5.2.1.1.1

Apply the distributive property.

$x \ln (1024) + \ln (16) = \ln (0.8) + 5 x \ln (5) + 2 \ln (5)$

Step 5.2.1.1.2

Simplify $5 \ln (5)$ by moving $5$ inside the logarithm.

$x \ln (1024) + \ln (16) = \ln (0.8) + x \ln (5^{5}) + 2 \ln (5)$

Step 5.2.1.1.3

Simplify $2 \ln (5)$ by moving $2$ inside the logarithm.

$x \ln (1024) + \ln (16) = \ln (0.8) + x \ln (5^{5}) + \ln (5^{2})$

Step 5.2.1.1.4

Simplify each term.

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Step 5.2.1.1.4.1

Raise $5$ to the power of $5$ .

$x \ln (1024) + \ln (16) = \ln (0.8) + x \ln (3125) + \ln (5^{2})$

Step 5.2.1.1.4.2

Raise $5$ to the power of $2$ .

$x \ln (1024) + \ln (16) = \ln (0.8) + x \ln (3125) + \ln (25)$

Step 5.2.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$x \ln (1024) + \ln (16) = x \ln (3125) + \ln (0.8 \cdot 25)$

Step 5.2.1.3

Multiply $0.8$ by $25$ .

$x \ln (1024) + \ln (16) = x \ln (3125) + \ln (20)$

Step 5.3

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

$x \ln (1024) + \ln (16) - x \ln (3125) - \ln (20) = 0$

Step 5.4

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

$x \ln (1024) - x \ln (3125) + \ln (\frac{16}{20}) = 0$

Step 5.5

Cancel the common factor of $16$ and $20$ .

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Step 5.5.1

Factor $4$ out of $16$ .

$x \ln (1024) - x \ln (3125) + \ln (\frac{4 (4)}{20}) = 0$

Step 5.5.2

Cancel the common factors.

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Step 5.5.2.1

Factor $4$ out of $20$ .

$x \ln (1024) - x \ln (3125) + \ln (\frac{4 \cdot 4}{4 \cdot 5}) = 0$

Step 5.5.2.2

Cancel the common factor.

$x \ln (1024) - x \ln (3125) + \ln (\frac{4 \cdot 4}{4 \cdot 5}) = 0$

Step 5.5.2.3

Rewrite the expression.

$x \ln (1024) - x \ln (3125) + \ln (\frac{4}{5}) = 0$

Step 5.6

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

$x \ln (1024) - x \ln (3125) = - \ln (\frac{4}{5})$

Step 5.7

Factor $x$ out of $x \ln (1024) - x \ln (3125)$ .

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Step 5.7.1

Factor $x$ out of $x \ln (1024)$ .

$x (\ln (1024)) - x \ln (3125) = - \ln (\frac{4}{5})$

Step 5.7.2

Factor $x$ out of $- x \ln (3125)$ .

$x (\ln (1024)) + x (- 1 \ln (3125)) = - \ln (\frac{4}{5})$

Step 5.7.3

Factor $x$ out of $x (\ln (1024)) + x (- 1 \ln (3125))$ .

$x (\ln (1024) - 1 \ln (3125)) = - \ln (\frac{4}{5})$

Step 5.8

Rewrite $- 1 \ln (3125)$ as $- \ln (3125)$ .

$x (\ln (1024) - \ln (3125)) = - \ln (\frac{4}{5})$

Step 5.9

Divide each term in $x (\ln (1024) - \ln (3125)) = - \ln (\frac{4}{5})$ by $\ln (1024) - \ln (3125)$ and simplify.

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Step 5.9.1

Divide each term in $x (\ln (1024) - \ln (3125)) = - \ln (\frac{4}{5})$ by $\ln (1024) - \ln (3125)$ .

$\frac{x (\ln (1024) - \ln (3125))}{\ln (1024) - \ln (3125)} = \frac{- \ln (\frac{4}{5})}{\ln (1024) - \ln (3125)}$

Step 5.9.2

Simplify the left side.

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Step 5.9.2.1

Cancel the common factor of $\ln (1024) - \ln (3125)$ .

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Step 5.9.2.1.1

Cancel the common factor.

$\frac{x (\ln (1024) - \ln (3125))}{\ln (1024) - \ln (3125)} = \frac{- \ln (\frac{4}{5})}{\ln (1024) - \ln (3125)}$

Step 5.9.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{4}{5})}{\ln (1024) - \ln (3125)}$

Step 5.9.3

Simplify the right side.

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Step 5.9.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{4}{5})}{\ln (1024) - \ln (3125)}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{\ln (\frac{4}{5})}{\ln (1024) - \ln (3125)}$

Decimal Form:

$x = - 0.2$