Finite Math Examples

$\log_{16} (x^{2} + 3 x) = \frac{1}{2}$

Step 1

Rewrite $\log_{16} (x^{2} + 3 x) = \frac{1}{2}$ 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 $b^{y} = x$ .

$16^{\frac{1}{2}} = x^{2} + 3 x$

Step 2

Solve for $x$ .

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Step 2.1

Rewrite the equation as $x^{2} + 3 x = 16^{\frac{1}{2}}$ .

$x^{2} + 3 x = 16^{\frac{1}{2}}$

Step 2.2

Simplify $16^{\frac{1}{2}}$ .

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Rewrite $16$ as $4^{2}$ .

$x^{2} + 3 x = {(4^{2})}^{\frac{1}{2}}$

Step 2.2.1.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{2} + 3 x = 4^{2 (\frac{1}{2})}$

Step 2.2.2

Cancel the common factor of $2$ .

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Step 2.2.2.1

Cancel the common factor.

$x^{2} + 3 x = 4^{2 (\frac{1}{2})}$

Step 2.2.2.2

Rewrite the expression.

$x^{2} + 3 x = 4^{1}$

Step 2.2.3

Evaluate the exponent.

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

Step 2.3

Subtract $4$ from both sides of the equation.

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

Step 2.4

Factor $x^{2} + 3 x - 4$ using the AC method.

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Step 2.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $3$ .

$- 1, 4$

Step 2.4.2

Write the factored form using these integers.

$(x - 1) (x + 4) = 0$

Step 2.5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x + 4 = 0$

Step 2.6

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.7

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.7.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.8

The final solution is all the values that make $(x - 1) (x + 4) = 0$ true.

$x = 1, - 4$