Precalculus Examples

$\ln (x) + \ln (x + 10) = 5$

Step 1

Simplify the left side.

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Step 1.1

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

$\ln (x (x + 10)) = 5$

Step 1.2

Apply the distributive property.

$\ln (x \cdot x + x \cdot 10) = 5$

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply $x$ by $x$ .

$\ln (x^{2} + x \cdot 10) = 5$

Step 1.3.2

Move $10$ to the left of $x$ .

$\ln (x^{2} + 10 x) = 5$

Step 2

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

$e^{\ln (x^{2} + 10 x)} = e^{5}$

Step 3

Rewrite $\ln (x^{2} + 10 x) = 5$ 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 $b^{y} = x$ .

$e^{5} = x^{2} + 10 x$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $x^{2} + 10 x = e^{5}$ .

$x^{2} + 10 x = e^{5}$

Step 4.2

Subtract $e^{5}$ from both sides of the equation.

$x^{2} + 10 x - e^{5} = 0$

Step 4.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.4

Substitute the values $a = 1$ , $b = 10$ , and $c = - e^{5}$ into the quadratic formula and solve for $x$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (1 \cdot (- e^{5}))}}{2 \cdot 1}$

Step 4.5

Simplify.

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Step 4.5.1

Simplify the numerator.

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Step 4.5.1.1

Raise $10$ to the power of $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot (- 1 e^{5})}}{2 \cdot 1}$

Step 4.5.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 4.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot (- 1 e^{5})}}{2 \cdot 1}$

Step 4.5.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 10 \pm \sqrt{100 + 4 e^{5}}}{2 \cdot 1}$

Step 4.5.1.3

Rewrite $100 + 4 e^{5}$ as $2^{2} (25 + e^{5})$ .

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Step 4.5.1.3.1

Factor $4$ out of $100$ .

$x = \frac{- 10 \pm \sqrt{4 (25) + 4 e^{5}}}{2 \cdot 1}$

Step 4.5.1.3.2

Factor $4$ out of $4 e^{5}$ .

$x = \frac{- 10 \pm \sqrt{4 (25) + 4 (e^{5})}}{2 \cdot 1}$

Step 4.5.1.3.3

Factor $4$ out of $4 (25) + 4 (e^{5})$ .

$x = \frac{- 10 \pm \sqrt{4 (25 + e^{5})}}{2 \cdot 1}$

Step 4.5.1.3.4

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 10 \pm \sqrt{2^{2} (25 + e^{5})}}{2 \cdot 1}$

Step 4.5.1.4

Pull terms out from under the radical.

$x = \frac{- 10 \pm 2 \sqrt{25 + e^{5}}}{2 \cdot 1}$

Step 4.5.2

Multiply $2$ by $1$ .

$x = \frac{- 10 \pm 2 \sqrt{25 + e^{5}}}{2}$

Step 4.5.3

Simplify $\frac{- 10 \pm 2 \sqrt{25 + e^{5}}}{2}$ .

$x = - 5 \pm \sqrt{25 + e^{5}}$

Step 4.6

The final answer is the combination of both solutions.

$x = - 5 + \sqrt{25 + e^{5}}, - 5 - \sqrt{25 + e^{5}}$

Step 5

Exclude the solutions that do not make $\ln (x) + \ln (x + 10) = 5$ true.

$x = - 5 + \sqrt{25 + e^{5}}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - 5 + \sqrt{25 + e^{5}}$

Decimal Form:

$x = 8.16864302 \dots$