Algebra Examples

$5^{x - 2} = 3 \cdot 4^{x + 1}$

Step 1

Take the log of both sides of the equation.

$\ln (5^{x - 2}) = \ln (3 \cdot 4^{x + 1})$

Step 2

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

$(x - 2) \ln (5) = \ln (3 \cdot 4^{x + 1})$

Step 3

Rewrite $\ln (3 \cdot 4^{x + 1})$ as $\ln (3) + \ln (4^{x + 1})$ .

$(x - 2) \ln (5) = \ln (3) + \ln (4^{x + 1})$

Step 4

Expand $\ln (4^{x + 1})$ by moving $x + 1$ outside the logarithm.

$(x - 2) \ln (5) = \ln (3) + (x + 1) \ln (4)$

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 $(x - 2) \ln (5)$ .

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Step 5.1.1.1

Apply the distributive property.

$x \ln (5) - 2 \ln (5) = \ln (3) + (x + 1) \ln (4)$

Step 5.1.1.2

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

$x \ln (5) - \ln (5^{2}) = \ln (3) + (x + 1) \ln (4)$

Step 5.1.1.3

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

$x \ln (5) - \ln (25) = \ln (3) + (x + 1) \ln (4)$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $\ln (3) + (x + 1) \ln (4)$ .

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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.

$x \ln (5) - \ln (25) = \ln (3) + x \ln (4) + 1 \ln (4)$

Step 5.2.1.1.2

Multiply $\ln (4)$ by $1$ .

$x \ln (5) - \ln (25) = \ln (3) + x \ln (4) + \ln (4)$

Step 5.2.1.2

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

$x \ln (5) - \ln (25) = x \ln (4) + \ln (3 \cdot 4)$

Step 5.2.1.3

Multiply $3$ by $4$ .

$x \ln (5) - \ln (25) = x \ln (4) + \ln (12)$

Step 5.3

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

$x \ln (5) - \ln (25) - x \ln (4) - \ln (12) = 0$

Step 5.4

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

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Step 5.4.1

Add $\ln (25)$ to both sides of the equation.

$x \ln (5) - x \ln (4) - \ln (12) = \ln (25)$

Step 5.4.2

Add $\ln (12)$ to both sides of the equation.

$x \ln (5) - x \ln (4) = \ln (25) + \ln (12)$

Step 5.5

Factor $x$ out of $x \ln (5) - x \ln (4)$ .

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Step 5.5.1

Factor $x$ out of $x \ln (5)$ .

$x (\ln (5)) - x \ln (4) = \ln (25) + \ln (12)$

Step 5.5.2

Factor $x$ out of $- x \ln (4)$ .

$x (\ln (5)) + x (- 1 \ln (4)) = \ln (25) + \ln (12)$

Step 5.5.3

Factor $x$ out of $x (\ln (5)) + x (- 1 \ln (4))$ .

$x (\ln (5) - 1 \ln (4)) = \ln (25) + \ln (12)$

Step 5.6

Rewrite $- 1 \ln (4)$ as $- \ln (4)$ .

$x (\ln (5) - \ln (4)) = \ln (25) + \ln (12)$

Step 5.7

Divide each term in $x (\ln (5) - \ln (4)) = \ln (25) + \ln (12)$ by $\ln (5) - \ln (4)$ and simplify.

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Step 5.7.1

Divide each term in $x (\ln (5) - \ln (4)) = \ln (25) + \ln (12)$ by $\ln (5) - \ln (4)$ .

$\frac{x (\ln (5) - \ln (4))}{\ln (5) - \ln (4)} = \frac{\ln (25)}{\ln (5) - \ln (4)} + \frac{\ln (12)}{\ln (5) - \ln (4)}$

Step 5.7.2

Simplify the left side.

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Step 5.7.2.1

Cancel the common factor of $\ln (5) - \ln (4)$ .

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Step 5.7.2.1.1

Cancel the common factor.

$\frac{x (\ln (5) - \ln (4))}{\ln (5) - \ln (4)} = \frac{\ln (25)}{\ln (5) - \ln (4)} + \frac{\ln (12)}{\ln (5) - \ln (4)}$

Step 5.7.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (25)}{\ln (5) - \ln (4)} + \frac{\ln (12)}{\ln (5) - \ln (4)}$

Step 5.7.3

Simplify the right side.

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Step 5.7.3.1

Combine the numerators over the common denominator.

$x = \frac{\ln (25) + \ln (12)}{\ln (5) - \ln (4)}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (25) + \ln (12)}{\ln (5) - \ln (4)}$

Decimal Form:

$x = 25.56104552 \dots$