Precalculus Examples

$7^{x^{2} - 48} = 49^{x}$

Step 1

Create equivalent expressions in the equation that all have equal bases.

$7^{x^{2} - 48} = 7^{2 x}$

Step 2

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

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

Step 3

Solve for $x$ .

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Step 3.1

Subtract $2 x$ from both sides of the equation.

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

Step 3.2

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

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Step 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 $- 2$ .

$- 8, 6$

Step 3.2.2

Write the factored form using these integers.

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

Step 3.3

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

$x - 8 = 0$

$x + 6 = 0$

Step 3.4

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

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Step 3.4.1

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

$x - 8 = 0$

Step 3.4.2

Add $8$ to both sides of the equation.

$x = 8$

Step 3.5

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

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Step 3.5.1

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

$x + 6 = 0$

Step 3.5.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 3.6

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

$x = 8, - 6$