Solve for x 3^(x+2)+3^(1-x)>28

$3^{x + 2} + 3^{1 - x} > 28$

Step 1

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

$3^{x} \cdot 3^{2} + 3^{1 - x} = 28$

Step 2

Rewrite $3^{1 - x}$ as $3^{1} \cdot 3^{- x}$ .

$3^{x} \cdot 3^{2} + 3 \cdot 3^{- x} = 28$

Step 3

Rewrite $3^{- x}$ as exponentiation.

$3^{x} \cdot 3^{2} + 3 \cdot {(3^{x})}^{- 1} = 28$

Step 4

Substitute $u$ for $3^{x}$ .

$u \cdot 3^{2} + 3 \cdot u^{- 1} = 28$

Step 5

Simplify each term.

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

$u \cdot 9 + 3 \cdot u^{- 1} = 28$

Move $9$ to the left of $u$ .

$9 u + 3 \cdot u^{- 1} = 28$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$9 u + 3 \cdot \frac{1}{u} = 28$

Combine $3$ and $\frac{1}{u}$ .

$9 u + \frac{3}{u} = 28$

Step 6

Solve for $u$ .

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, u, 1$

The LCM of one and any expression is the expression.

$u$

Multiply each term in $9 u + \frac{3}{u} = 28$ by $u$ to eliminate the fractions.

Multiply each term in $9 u + \frac{3}{u} = 28$ by $u$ .

$9 u \cdot u + \frac{3}{u} u = 28 u$

Simplify the left side.

Simplify each term.

Multiply $u$ by $u$ by adding the exponents.

Move $u$ .

$9 (u \cdot u) + \frac{3}{u} u = 28 u$

Multiply $u$ by $u$ .

$9 u^{2} + \frac{3}{u} u = 28 u$

Cancel the common factor of $u$ .

Cancel the common factor.

$9 u^{2} + \frac{3}{u} u = 28 u$

Rewrite the expression.

$9 u^{2} + 3 = 28 u$

Solve the equation.

Subtract $28 u$ from both sides of the equation.

$9 u^{2} + 3 - 28 u = 0$

Factor by grouping.

Reorder terms.

$9 u^{2} - 28 u + 3 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot 3 = 27$ and whose sum is $b = - 28$ .

Factor $- 28$ out of $- 28 u$ .

$9 u^{2} - 28 u + 3 = 0$

Rewrite $- 28$ as $- 1$ plus $- 27$

$9 u^{2} + (- 1 - 27) u + 3 = 0$

Apply the distributive property.

$9 u^{2} - 1 u - 27 u + 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(9 u^{2} - 1 u) - 27 u + 3 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (9 u - 1) - 3 (9 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $9 u - 1$ .

$(9 u - 1) (u - 3) = 0$

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

$9 u - 1 = 0$

$u - 3 = 0$

Set $9 u - 1$ equal to $0$ and solve for $u$ .

Set $9 u - 1$ equal to $0$ .

$9 u - 1 = 0$

Solve $9 u - 1 = 0$ for $u$ .

Add $1$ to both sides of the equation.

$9 u = 1$

Divide each term in $9 u = 1$ by $9$ and simplify.

Divide each term in $9 u = 1$ by $9$ .

$\frac{9 u}{9} = \frac{1}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 u}{9} = \frac{1}{9}$

Divide $u$ by $1$ .

$u = \frac{1}{9}$

Set $u - 3$ equal to $0$ and solve for $u$ .

Set $u - 3$ equal to $0$ .

$u - 3 = 0$

Add $3$ to both sides of the equation.

$u = 3$

The final solution is all the values that make $(9 u - 1) (u - 3) = 0$ true.

$u = \frac{1}{9}, 3$

Step 7

Substitute $\frac{1}{9}$ for $u$ in $u = 3^{x}$ .

$\frac{1}{9} = 3^{x}$

Step 8

Solve $\frac{1}{9} = 3^{x}$ .

Rewrite the equation as $3^{x} = \frac{1}{9}$ .

$3^{x} = \frac{1}{9}$

Raise $9$ to the power of $1$ .

$3^{x} = \frac{1}{9^{1}}$

Move $9^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$3^{x} = 9^{- 1}$

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

$3^{x} = 3^{- 2}$

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

$x = - 2$

Step 9

Substitute $3$ for $u$ in $u = 3^{x}$ .

$3 = 3^{x}$

Step 10

Solve $3 = 3^{x}$ .

Rewrite the equation as $3^{x} = 3$ .

$3^{x} = 3$

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

$3^{x} = 3^{1}$

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

$x = 1$

Step 11

List the solutions that makes the equation true.

$x = - 2, 1$

Step 12

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$x > 1$

Step 13

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$3^{(- 4) + 2} + 3^{1 - (- 4)} > 28$

The left side $243.\overline{1}$ is greater than the right side $28$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$3^{(0) + 2} + 3^{1 - (0)} > 28$

The left side $12$ is not greater than the right side $28$ , which means that the given statement is false.

False

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$3^{(4) + 2} + 3^{1 - (4)} > 28$

The left side $729.\overline{037}$ is greater than the right side $28$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 2$ True

$- 2 < x < 1$ False

$x > 1$ True

$x < - 2$ True

$- 2 < x < 1$ False

$x > 1$ True

Step 14

The solution consists of all of the true intervals.

$x < - 2$ or $x > 1$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$x < - 2 or x > 1$

Interval Notation:

$(- \infty, - 2) \cup (1, \infty)$

Step 16

Solve for x 3^(x+2)+3^(1-x)>28

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