Basic Math Examples

$3 \cdot 12^{q} = 2^{q - 1}$

Step 1

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

$\ln (3 \cdot 12^{q}) = \ln (2^{q - 1})$

Step 2

Expand the left side.

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Step 2.1

Rewrite $\ln (3 \cdot 12^{q})$ as $\ln (3) + \ln (12^{q})$ .

$\ln (3) + \ln (12^{q}) = \ln (2^{q - 1})$

Step 2.2

Expand $\ln (12^{q})$ by moving $q$ outside the logarithm.

$\ln (3) + q \ln (12) = \ln (2^{q - 1})$

Step 3

Expand $\ln (2^{q - 1})$ by moving $q - 1$ outside the logarithm.

$\ln (3) + q \ln (12) = (q - 1) \ln (2)$

Step 4

Simplify the right side.

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Step 4.1

Simplify $(q - 1) \ln (2)$ .

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Step 4.1.1

Apply the distributive property.

$\ln (3) + q \ln (12) = q \ln (2) - 1 \ln (2)$

Step 4.1.2

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

$\ln (3) + q \ln (12) = q \ln (2) - \ln (2)$

Step 5

Reorder $\ln (3)$ and $q \ln (12)$ .

$q \ln (12) + \ln (3) = q \ln (2) - \ln (2)$

Step 6

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

$q \ln (12) + \ln (3) - q \ln (2) + \ln (2) = 0$

Step 7

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

$q \ln (12) - q \ln (2) + \ln (3 \cdot 2) = 0$

Step 8

Multiply $3$ by $2$ .

$q \ln (12) - q \ln (2) + \ln (6) = 0$

Step 9

Subtract $\ln (6)$ from both sides of the equation.

$q \ln (12) - q \ln (2) = - \ln (6)$

Step 10

Factor $q$ out of $q \ln (12) - q \ln (2)$ .

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Step 10.1

Factor $q$ out of $q \ln (12)$ .

$q (\ln (12)) - q \ln (2) = - \ln (6)$

Step 10.2

Factor $q$ out of $- q \ln (2)$ .

$q (\ln (12)) + q (- 1 \ln (2)) = - \ln (6)$

Step 10.3

Factor $q$ out of $q (\ln (12)) + q (- 1 \ln (2))$ .

$q (\ln (12) - 1 \ln (2)) = - \ln (6)$

Step 11

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

$q (\ln (12) - \ln (2)) = - \ln (6)$

Step 12

Divide each term in $q (\ln (12) - \ln (2)) = - \ln (6)$ by $\ln (12) - \ln (2)$ and simplify.

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Step 12.1

Divide each term in $q (\ln (12) - \ln (2)) = - \ln (6)$ by $\ln (12) - \ln (2)$ .

$\frac{q (\ln (12) - \ln (2))}{\ln (12) - \ln (2)} = \frac{- \ln (6)}{\ln (12) - \ln (2)}$

Step 12.2

Simplify the left side.

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Step 12.2.1

Cancel the common factor of $\ln (12) - \ln (2)$ .

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Step 12.2.1.1

Cancel the common factor.

$\frac{q (\ln (12) - \ln (2))}{\ln (12) - \ln (2)} = \frac{- \ln (6)}{\ln (12) - \ln (2)}$

Step 12.2.1.2

Divide $q$ by $1$ .

$q = \frac{- \ln (6)}{\ln (12) - \ln (2)}$

Step 12.3

Simplify the right side.

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Step 12.3.1

Move the negative in front of the fraction.

$q = - \frac{\ln (6)}{\ln (12) - \ln (2)}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$q = - \frac{\ln (6)}{\ln (12) - \ln (2)}$

Decimal Form:

$q = - 1$