Algebra Examples

$2^{2 + x} \cdot 3^{x} = \frac{2}{3} \cdot 6^{3 x - 1}$

Step 1

Take the log of both sides of the equation.

$\ln (2^{2 + x} \cdot 3^{x}) = \ln (\frac{2}{3} \cdot 6^{3 x - 1})$

Step 2

Rewrite $\ln (2^{2 + x} \cdot 3^{x})$ as $\ln (2^{2 + x}) + \ln (3^{x})$ .

$\ln (2^{2 + x}) + \ln (3^{x}) = \ln (\frac{2}{3} \cdot 6^{3 x - 1})$

Step 3

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

$(2 + x) \ln (2) + \ln (3^{x}) = \ln (\frac{2}{3} \cdot 6^{3 x - 1})$

Step 4

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

$(2 + x) \ln (2) + x \ln (3) = \ln (\frac{2}{3} \cdot 6^{3 x - 1})$

Step 5

Rewrite $\ln (\frac{2}{3} \cdot 6^{3 x - 1})$ as $\ln (\frac{2}{3}) + \ln (6^{3 x - 1})$ .

$(2 + x) \ln (2) + x \ln (3) = \ln (\frac{2}{3}) + \ln (6^{3 x - 1})$

Step 6

Rewrite $\ln (\frac{2}{3})$ as $\ln (2) - \ln (3)$ .

$(2 + x) \ln (2) + x \ln (3) = \ln (2) - \ln (3) + \ln (6^{3 x - 1})$

Step 7

Expand $\ln (6^{3 x - 1})$ by moving $3 x - 1$ outside the logarithm.

$(2 + x) \ln (2) + x \ln (3) = \ln (2) - \ln (3) + (3 x - 1) \ln (6)$

Step 8

Solve the equation for $x$ .

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Step 8.1

Simplify the left side.

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Step 8.1.1

Apply the distributive property.

$2 \ln (2) + x \ln (2) + x \ln (3) = \ln (2) - \ln (3) + (3 x - 1) \ln (6)$

Step 8.1.2

Move $2 \ln (2)$ .

$x \ln (2) + x \ln (3) + 2 \ln (2) = \ln (2) - \ln (3) + (3 x - 1) \ln (6)$

Step 8.2

Simplify the right side.

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Step 8.2.1

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

$x \ln (2) + x \ln (3) + 2 \ln (2) = \ln (\frac{2}{3}) + (3 x - 1) \ln (6)$

Step 8.2.2

Simplify each term.

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Step 8.2.2.1

Apply the distributive property.

$x \ln (2) + x \ln (3) + 2 \ln (2) = \ln (\frac{2}{3}) + 3 x \ln (6) - 1 \ln (6)$

Step 8.2.2.2

Rewrite $- 1 \ln (6)$ as $- \ln (6)$ .

$x \ln (2) + x \ln (3) + 2 \ln (2) = \ln (\frac{2}{3}) + 3 x \ln (6) - \ln (6)$

Step 8.2.3

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

$x \ln (2) + x \ln (3) + 2 \ln (2) = 3 x \ln (6) + \ln (\frac{\frac{2}{3}}{6})$

Step 8.2.4

Simplify each term.

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Step 8.2.4.1

Multiply the numerator by the reciprocal of the denominator.

$x \ln (2) + x \ln (3) + 2 \ln (2) = 3 x \ln (6) + \ln (\frac{2}{3} \cdot \frac{1}{6})$

Step 8.2.4.2

Cancel the common factor of $2$ .

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Step 8.2.4.2.1

Factor $2$ out of $6$ .

$x \ln (2) + x \ln (3) + 2 \ln (2) = 3 x \ln (6) + \ln (\frac{2}{3} \cdot \frac{1}{2 (3)})$

Step 8.2.4.2.2

Cancel the common factor.

$x \ln (2) + x \ln (3) + 2 \ln (2) = 3 x \ln (6) + \ln (\frac{2}{3} \cdot \frac{1}{2 \cdot 3})$

Step 8.2.4.2.3

Rewrite the expression.

$x \ln (2) + x \ln (3) + 2 \ln (2) = 3 x \ln (6) + \ln (\frac{1}{3} \cdot \frac{1}{3})$

Step 8.2.4.3

Multiply $\frac{1}{3}$ by $\frac{1}{3}$ .

$x \ln (2) + x \ln (3) + 2 \ln (2) = 3 x \ln (6) + \ln (\frac{1}{3 \cdot 3})$

Step 8.2.4.4

Multiply $3$ by $3$ .

$x \ln (2) + x \ln (3) + 2 \ln (2) = 3 x \ln (6) + \ln (\frac{1}{9})$

Step 8.3

Simplify the left side.

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Step 8.3.1

Simplify each term.

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Step 8.3.1.1

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

$x \ln (2) + x \ln (3) + \ln (2^{2}) = 3 x \ln (6) + \ln (\frac{1}{9})$

Step 8.3.1.2

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

$x \ln (2) + x \ln (3) + \ln (4) = 3 x \ln (6) + \ln (\frac{1}{9})$

Step 8.4

Simplify the right side.

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Step 8.4.1

Simplify each term.

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Step 8.4.1.1

Simplify $3 \ln (6)$ by moving $3$ inside the logarithm.

$x \ln (2) + x \ln (3) + \ln (4) = x \ln (6^{3}) + \ln (\frac{1}{9})$

Step 8.4.1.2

Raise $6$ to the power of $3$ .

$x \ln (2) + x \ln (3) + \ln (4) = x \ln (216) + \ln (\frac{1}{9})$

Step 8.5

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

$x \ln (2) + x \ln (3) + \ln (4) - x \ln (216) - \ln (\frac{1}{9}) = 0$

Step 8.6

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

$x \ln (2) + x \ln (3) - x \ln (216) + \ln (\frac{4}{\frac{1}{9}}) = 0$

Step 8.7

Simplify each term.

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Step 8.7.1

Multiply the numerator by the reciprocal of the denominator.

$x \ln (2) + x \ln (3) - x \ln (216) + \ln (4 \cdot 9) = 0$

Step 8.7.2

Multiply $4$ by $9$ .

$x \ln (2) + x \ln (3) - x \ln (216) + \ln (36) = 0$

Step 8.8

Subtract $\ln (36)$ from both sides of the equation.

$x \ln (2) + x \ln (3) - x \ln (216) = - \ln (36)$

Step 8.9

Factor $x$ out of $x \ln (2) + x \ln (3) - x \ln (216)$ .

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Step 8.9.1

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

$x (\ln (2)) + x \ln (3) - x \ln (216) = - \ln (36)$

Step 8.9.2

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

$x (\ln (2)) + x (\ln (3)) - x \ln (216) = - \ln (36)$

Step 8.9.3

Factor $x$ out of $- x \ln (216)$ .

$x (\ln (2)) + x (\ln (3)) + x (- 1 \ln (216)) = - \ln (36)$

Step 8.9.4

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

$x (\ln (2) + \ln (3)) + x (- 1 \ln (216)) = - \ln (36)$

Step 8.9.5

Factor $x$ out of $x (\ln (2) + \ln (3)) + x (- 1 \ln (216))$ .

$x (\ln (2) + \ln (3) - 1 \ln (216)) = - \ln (36)$

Step 8.10

Rewrite $- 1 \ln (216)$ as $- \ln (216)$ .

$x (\ln (2) + \ln (3) - \ln (216)) = - \ln (36)$

Step 8.11

Divide each term in $x (\ln (2) + \ln (3) - \ln (216)) = - \ln (36)$ by $\ln (2) + \ln (3) - \ln (216)$ and simplify.

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Step 8.11.1

Divide each term in $x (\ln (2) + \ln (3) - \ln (216)) = - \ln (36)$ by $\ln (2) + \ln (3) - \ln (216)$ .

$\frac{x (\ln (2) + \ln (3) - \ln (216))}{\ln (2) + \ln (3) - \ln (216)} = \frac{- \ln (36)}{\ln (2) + \ln (3) - \ln (216)}$

Step 8.11.2

Simplify the left side.

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Step 8.11.2.1

Cancel the common factor of $\ln (2) + \ln (3) - \ln (216)$ .

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Step 8.11.2.1.1

Cancel the common factor.

$\frac{x (\ln (2) + \ln (3) - \ln (216))}{\ln (2) + \ln (3) - \ln (216)} = \frac{- \ln (36)}{\ln (2) + \ln (3) - \ln (216)}$

Step 8.11.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (36)}{\ln (2) + \ln (3) - \ln (216)}$

Step 8.11.3

Simplify the right side.

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Step 8.11.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (36)}{\ln (2) + \ln (3) - \ln (216)}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{\ln (36)}{\ln (2) + \ln (3) - \ln (216)}$

Decimal Form:

$x = 1$