Algebra Examples

$2^{x} \cdot 3^{x + 1} = 15$

Step 1

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

$\ln (2^{x} \cdot 3^{x + 1}) = \ln (15)$

Step 2

Expand the left side.

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Step 2.1

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

$\ln (2^{x}) + \ln (3^{x + 1}) = \ln (15)$

Step 2.2

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

$x \ln (2) + \ln (3^{x + 1}) = \ln (15)$

Step 2.3

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

$x \ln (2) + (x + 1) \ln (3) = \ln (15)$

Step 3

Simplify the left side.

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Step 3.1

Simplify each term.

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Step 3.1.1

Apply the distributive property.

$x \ln (2) + x \ln (3) + 1 \ln (3) = \ln (15)$

Step 3.1.2

Multiply $\ln (3)$ by $1$ .

$x \ln (2) + x \ln (3) + \ln (3) = \ln (15)$

Step 4

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

$x \ln (2) + x \ln (3) + \ln (3) - \ln (15) = 0$

Step 5

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

$x \ln (2) + x \ln (3) + \ln (\frac{3}{15}) = 0$

Step 6

Cancel the common factor of $3$ and $15$ .

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Step 6.1

Factor $3$ out of $3$ .

$x \ln (2) + x \ln (3) + \ln (\frac{3 (1)}{15}) = 0$

Step 6.2

Cancel the common factors.

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Step 6.2.1

Factor $3$ out of $15$ .

$x \ln (2) + x \ln (3) + \ln (\frac{3 \cdot 1}{3 \cdot 5}) = 0$

Step 6.2.2

Cancel the common factor.

$x \ln (2) + x \ln (3) + \ln (\frac{3 \cdot 1}{3 \cdot 5}) = 0$

Step 6.2.3

Rewrite the expression.

$x \ln (2) + x \ln (3) + \ln (\frac{1}{5}) = 0$

Step 7

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

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

Step 8

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

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Step 8.1

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

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

Step 8.2

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

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

Step 8.3

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

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

Step 9

Divide each term in $x (\ln (2) + \ln (3)) = - \ln (\frac{1}{5})$ by $\ln (2) + \ln (3)$ and simplify.

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Step 9.1

Divide each term in $x (\ln (2) + \ln (3)) = - \ln (\frac{1}{5})$ by $\ln (2) + \ln (3)$ .

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

Step 9.2

Simplify the left side.

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Step 9.2.1

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

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Step 9.2.1.1

Cancel the common factor.

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

Step 9.2.1.2

Divide $x$ by $1$ .

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

Step 9.3

Simplify the right side.

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Step 9.3.1

Move the negative in front of the fraction.

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

Step 10

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.89824440 \dots$