Precalculus Examples

$\log_{g} (2 x) = \log_{g} (48 - x^{2})$

Step 1

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$2 x = 48 - x^{2}$

Step 2

Solve for $x$ .

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Step 2.1

Add $x^{2}$ to both sides of the equation.

$2 x + x^{2} = 48$

Step 2.2

Subtract $48$ from both sides of the equation.

$2 x + x^{2} - 48 = 0$

Step 2.3

Factor the left side of the equation.

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Step 2.3.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$2 u + u^{2} - 48 = 0$

Step 2.3.2

Factor $2 u + u^{2} - 48$ using the AC method.

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Step 2.3.2.1

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 $- 48$ and whose sum is $2$ .

$- 6, 8$

Step 2.3.2.2

Write the factored form using these integers.

$(u - 6) (u + 8) = 0$

Step 2.3.3

Replace all occurrences of $u$ with $x$ .

$(x - 6) (x + 8) = 0$

Step 2.4

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

$x - 6 = 0$

$x + 8 = 0$

Step 2.5

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.5.2

Add $6$ to both sides of the equation.

$x = 6$

Step 2.6

Set $x + 8$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 2.6.2

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 2.7

The final solution is all the values that make $(x - 6) (x + 8) = 0$ true.

$x = 6, - 8$