Solve for x 8^(x^2-14/3x)=32

$8^{x^{2} - \frac{14}{3} x} = 32$

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

$2^{3 (x^{2} - \frac{14}{3} x)} = 2^{5}$

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

$3 (x^{2} - \frac{14}{3} x) = 5$

Solve for $x$ .

Simplify.

Simplify each term.

Combine $x$ and $\frac{14}{3}$ .

$3 (x^{2} - \frac{x \cdot 14}{3}) = 5$

Move $14$ to the left of $x$ .

$3 (x^{2} - \frac{14 x}{3}) = 5$

Apply the distributive property.

$3 x^{2} + 3 (- \frac{14 x}{3}) = 5$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{14 x}{3}$ into the numerator.

$3 x^{2} + 3 \frac{- 14 x}{3} = 5$

Cancel the common factor.

$3 x^{2} + 3 \frac{- 14 x}{3} = 5$

Rewrite the expression.

$3 x^{2} - 14 x = 5$

Divide each term in $(3 x^{2} - 14 x) = 5$ by $3$ and simplify.

Divide each term in $(3 x^{2} - 14 x) = 5$ by $3$ .

$\frac{(3 x^{2} - 14 x)}{3} = \frac{5}{3}$

Simplify the left side.

Factor $x$ out of $3 x^{2} - 14 x$ .

Factor $x$ out of $3 x^{2}$ .

$\frac{x (3 x) - 14 x}{3} = \frac{5}{3}$

Factor $x$ out of $- 14 x$ .

$\frac{x (3 x) + x \cdot - 14}{3} = \frac{5}{3}$

Factor $x$ out of $x (3 x) + x \cdot - 14$ .

$\frac{x (3 x - 14)}{3} = \frac{5}{3}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x (3 x - 14) = 5$

Simplify $x (3 x - 14)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (3 x) + x \cdot - 14 = 5$

Reorder.

Rewrite using the commutative property of multiplication.

$3 x \cdot x + x \cdot - 14 = 5$

Move $- 14$ to the left of $x$ .

$3 x \cdot x - 14 \cdot x = 5$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$3 (x \cdot x) - 14 \cdot x = 5$

Multiply $x$ by $x$ .

$3 x^{2} - 14 \cdot x = 5$

$3 x^{2} - 14 x = 5$

Subtract $5$ from both sides of the equation.

$3 x^{2} - 14 x - 5 = 0$

Factor by grouping.

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 $a \cdot c = 3 \cdot - 5 = - 15$ and whose sum is $b = - 14$ .

Factor $- 14$ out of $- 14 x$ .

$3 x^{2} - 14 x - 5 = 0$

Rewrite $- 14$ as $1$ plus $- 15$

$3 x^{2} + (1 - 15) x - 5 = 0$

Apply the distributive property.

$3 x^{2} + 1 x - 15 x - 5 = 0$

Multiply $x$ by $1$ .

$3 x^{2} + x - 15 x - 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} + x) - 15 x - 5 = 0$

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

$x (3 x + 1) - 5 (3 x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

$(3 x + 1) (x - 5) = 0$

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

$3 x + 1 = 0$

$x - 5 = 0$

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $(3 x + 1) (x - 5) = 0$ true.

$x = - \frac{1}{3}, 5$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{3}, 5$

Decimal Form:

$x = - 0.\overline{3}, 5$

Solve for x 8^(x^2-14/3x)=32

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