Precalculus Examples

$\ln (A) \ln (B) = \ln (A) + \ln (B)$

Step 1

Solve the equation for $A$ .

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Step 1.1

Simplify the right side.

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Step 1.1.1

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

$\ln (A) \ln (B) = \ln (A B)$

Step 1.2

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

$\ln (A) \ln (B) - \ln (A B) = 0$

Step 1.3

To solve for $A$ , rewrite the equation using properties of logarithms.

$e^{\ln (A B)} = e^{\ln (A) \ln (B)}$

Step 1.4

Rewrite $\ln (A B) = \ln (A) \ln (B)$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{\ln (A) \ln (B)} = A B$

Step 1.5

Solve for $A$ .

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Step 1.5.1

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

$\ln (e^{\ln (A) \ln (B)}) = \ln (A B)$

Step 1.5.2

Expand the left side.

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Step 1.5.2.1

Expand $\ln (e^{\ln (A) \ln (B)})$ by moving $\ln (A) \ln (B)$ outside the logarithm.

$\ln (A) \ln (B) \ln (e) = \ln (A B)$

Step 1.5.2.2

The natural logarithm of $e$ is $1$ .

$\ln (A) \ln (B) \cdot 1 = \ln (A B)$

Step 1.5.2.3

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

$\ln (A) \ln (B) = \ln (A B)$

Step 1.5.3

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

$\ln (A) \ln (B) - \ln (A B) = 0$

Step 1.5.4

To solve for $A$ , rewrite the equation using properties of logarithms.

$e^{\ln (A B)} = e^{\ln (A) \ln (B)}$

Step 1.5.5

$e^{\ln (A) \ln (B)} = A B$

Step 1.5.6

Solve for $A$ .

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Step 1.5.6.1

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

$\ln (e^{\ln (A) \ln (B)}) = \ln (A B)$

Step 1.5.6.2

Expand the left side.

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Step 1.5.6.2.1

Expand $\ln (e^{\ln (A) \ln (B)})$ by moving $\ln (A) \ln (B)$ outside the logarithm.

$\ln (A) \ln (B) \ln (e) = \ln (A B)$

Step 1.5.6.2.2

The natural logarithm of $e$ is $1$ .

$\ln (A) \ln (B) \cdot 1 = \ln (A B)$

Step 1.5.6.2.3

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

$\ln (A) \ln (B) = \ln (A B)$

Step 1.5.6.3

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

$\ln (A) \ln (B) - \ln (A B) = 0$

Step 1.5.6.4

To solve for $A$ , rewrite the equation using properties of logarithms.

$e^{\ln (A B)} = e^{\ln (A) \ln (B)}$

Step 1.5.6.5

$e^{\ln (A) \ln (B)} = A B$

Step 1.5.6.6

Solve for $A$ .

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Step 1.5.6.6.1

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

$\ln (e^{\ln (A) \ln (B)}) = \ln (A B)$

Step 1.5.6.6.2

Expand the left side.

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Step 1.5.6.6.2.1

Expand $\ln (e^{\ln (A) \ln (B)})$ by moving $\ln (A) \ln (B)$ outside the logarithm.

$\ln (A) \ln (B) \ln (e) = \ln (A B)$

Step 1.5.6.6.2.2

The natural logarithm of $e$ is $1$ .

$\ln (A) \ln (B) \cdot 1 = \ln (A B)$

Step 1.5.6.6.2.3

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

$\ln (A) \ln (B) = \ln (A B)$

Step 1.5.6.6.3

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

$\ln (A) \ln (B) - \ln (A B) = 0$

Step 1.5.6.6.4

To solve for $A$ , rewrite the equation using properties of logarithms.

$e^{\ln (A B)} = e^{\ln (A) \ln (B)}$

Step 1.5.6.6.5

$e^{\ln (A) \ln (B)} = A B$

Step 1.5.6.6.6

Solve for $A$ .

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Step 1.5.6.6.6.1

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

$\ln (e^{\ln (A) \ln (B)}) = \ln (A B)$

Step 1.5.6.6.6.2

Expand the left side.

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Step 1.5.6.6.6.2.1

Expand $\ln (e^{\ln (A) \ln (B)})$ by moving $\ln (A) \ln (B)$ outside the logarithm.

$\ln (A) \ln (B) \ln (e) = \ln (A B)$

Step 1.5.6.6.6.2.2

The natural logarithm of $e$ is $1$ .

$\ln (A) \ln (B) \cdot 1 = \ln (A B)$

Step 1.5.6.6.6.2.3

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

$\ln (A) \ln (B) = \ln (A B)$

Step 1.5.6.6.6.3

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

$\ln (A) \ln (B) - \ln (A B) = 0$

Step 1.5.6.6.6.4

To solve for $A$ , rewrite the equation using properties of logarithms.

$e^{\ln (A B)} = e^{\ln (A) \ln (B)}$

Step 1.5.6.6.6.5

$e^{\ln (A) \ln (B)} = A B$

Step 1.5.6.6.6.6

Solve for $A$ .

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Step 1.5.6.6.6.6.1

Subtract $A B$ from both sides of the equation.

$e^{\ln (A) \ln (B)} - A B = 0$

Step 1.5.6.6.6.6.2

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

$e^{\ln (A) \ln (B)} = A B + 0$

Step 1.5.6.6.6.6.3

Add $A B$ and $0$ .

$e^{\ln (A) \ln (B)} = A B$

Step 1.5.6.6.6.6.4

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

$\ln (e^{\ln (A) \ln (B)}) = \ln (A B)$

Step 1.5.6.6.6.6.5

Expand the left side.

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Step 1.5.6.6.6.6.5.1

Expand $\ln (e^{\ln (A) \ln (B)})$ by moving $\ln (A) \ln (B)$ outside the logarithm.

$\ln (A) \ln (B) \ln (e) = \ln (A B)$

Step 1.5.6.6.6.6.5.2

The natural logarithm of $e$ is $1$ .

$\ln (A) \ln (B) \cdot 1 = \ln (A B)$

Step 1.5.6.6.6.6.5.3

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

$\ln (A) \ln (B) = \ln (A B)$

Step 2

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be $0$ or $1$ for each of its variables. In this case, the degrees of the variables in the equation violate the linear equation definition, which means that the equation is not a linear equation.

Not Linear