Pre-Algebra Examples

$2^{X + 3} + 4^{X + 1} - 320 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite $2^{X + 3}$ as $2^{X} \cdot 2^{3}$ .

$2^{X} \cdot 2^{3} + 4^{X + 1} - 320 = 0$

Step 1.2

Rewrite $4^{X + 1}$ as $4^{X} \cdot 4^{1}$ .

$2^{X} \cdot 2^{3} + 4^{X} \cdot 4 - 320 = 0$

Step 1.3

Rewrite $2^{X}$ as ${(2^{2})}^{X}$ .

$2^{X} \cdot 2^{3} + {(2^{2})}^{X} \cdot 4 - 320 = 0$

Step 1.4

Rewrite ${(2^{2})}^{X}$ as ${(2^{X})}^{2}$ .

$2^{X} \cdot 2^{3} + {(2^{X})}^{2} \cdot 4 - 320 = 0$

Step 1.5

Let $u = 2^{X}$ . Substitute $u$ for all occurrences of $2^{X}$ .

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Step 1.5.1

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

$u \cdot 8 + u^{2} \cdot 4 - 320 = 0$

Step 1.5.2

Move $8$ to the left of $u$ .

$8 u + u^{2} \cdot 4 - 320 = 0$

Step 1.5.3

Move $4$ to the left of $u^{2}$ .

$8 u + 4 u^{2} - 320 = 0$

Step 1.6

Factor $4$ out of $8 u + 4 u^{2} - 320$ .

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Step 1.6.1

Factor $4$ out of $8 u$ .

$4 (2 u) + 4 u^{2} - 320 = 0$

Step 1.6.2

Factor $4$ out of $4 u^{2}$ .

$4 (2 u) + 4 (u^{2}) - 320 = 0$

Step 1.6.3

Factor $4$ out of $- 320$ .

$4 (2 u) + 4 u^{2} + 4 \cdot - 80 = 0$

Step 1.6.4

Factor $4$ out of $4 (2 u) + 4 u^{2}$ .

$4 (2 u + u^{2}) + 4 \cdot - 80 = 0$

Step 1.6.5

Factor $4$ out of $4 (2 u + u^{2}) + 4 \cdot - 80$ .

$4 (2 u + u^{2} - 80) = 0$

Step 1.7

Factor.

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Step 1.7.1

Factor $2 u + u^{2} - 80$ using the AC method.

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Step 1.7.1.1

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 $- 80$ and whose sum is $2$ .

$- 8, 10$

Step 1.7.1.2

Write the factored form using these integers.

$4 ((u - 8) (u + 10)) = 0$

Step 1.7.2

Remove unnecessary parentheses.

$4 (u - 8) (u + 10) = 0$

Step 1.8

Replace all occurrences of $u$ with $2^{X}$ .

$4 (2^{X} - 8) (2^{X} + 10) = 0$

Step 2

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

$2^{X} - 8 = 0$

$2^{X} + 10 = 0$

Step 3

Set $2^{X} - 8$ equal to $0$ and solve for $X$ .

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Step 3.1

Set $2^{X} - 8$ equal to $0$ .

$2^{X} - 8 = 0$

Step 3.2

Solve $2^{X} - 8 = 0$ for $X$ .

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Step 3.2.1

Add $8$ to both sides of the equation.

$2^{X} = 8$

Step 3.2.2

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

$2^{X} = 2^{3}$

Step 3.2.3

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

$X = 3$

Step 4

Set $2^{X} + 10$ equal to $0$ and solve for $X$ .

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Step 4.1

Set $2^{X} + 10$ equal to $0$ .

$2^{X} + 10 = 0$

Step 4.2

Solve $2^{X} + 10 = 0$ for $X$ .

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Step 4.2.1

Subtract $10$ from both sides of the equation.

$2^{X} = - 10$

Step 4.2.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (2^{X}) = \ln (- 10)$

Step 4.2.3

The equation cannot be solved because $\ln (- 10)$ is undefined.

Undefined

Step 4.2.4

There is no solution for $2^{X} = - 10$

No solution

Step 5

The final solution is all the values that make $4 (2^{X} - 8) (2^{X} + 10) = 0$ true.

$X = 3$