Precalculus Examples

$\log_{5} (x^{2} + 3 x + 4) = 2$

Step 1

Rewrite $\log_{5} (x^{2} + 3 x + 4) = 2$ 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 $b^{y} = x$ .

$5^{2} = x^{2} + 3 x + 4$

Step 2

Solve for $x$ .

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Step 2.1

Rewrite the equation as $x^{2} + 3 x + 4 = 5^{2}$ .

$x^{2} + 3 x + 4 = 5^{2}$

Step 2.2

Raise $5$ to the power of $2$ .

$x^{2} + 3 x + 4 = 25$

Step 2.3

Move all terms to the left side of the equation and simplify.

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Step 2.3.1

Subtract $25$ from both sides of the equation.

$x^{2} + 3 x + 4 - 25 = 0$

Step 2.3.2

Subtract $25$ from $4$ .

$x^{2} + 3 x - 21 = 0$

Step 2.4

Use the quadratic formula to find the solutions.

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

Step 2.5

Substitute the values $a = 1$ , $b = 3$ , and $c = - 21$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 21)}}{2 \cdot 1}$

Step 2.6

Simplify.

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Step 2.6.1

Simplify the numerator.

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Step 2.6.1.1

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 21}}{2 \cdot 1}$

Step 2.6.1.2

Multiply $- 4 \cdot 1 \cdot - 21$ .

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Step 2.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 21}}{2 \cdot 1}$

Step 2.6.1.2.2

Multiply $- 4$ by $- 21$ .

$x = \frac{- 3 \pm \sqrt{9 + 84}}{2 \cdot 1}$

Step 2.6.1.3

Add $9$ and $84$ .

$x = \frac{- 3 \pm \sqrt{93}}{2 \cdot 1}$

Step 2.6.2

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{93}}{2}$

Step 2.7

The final answer is the combination of both solutions.

$x = - \frac{3 - \sqrt{93}}{2}, - \frac{3 + \sqrt{93}}{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{3 - \sqrt{93}}{2}, - \frac{3 + \sqrt{93}}{2}$

Decimal Form:

$x = 3.32182538 \dots, - 6.32182538 \dots$