Solve for x 2401^(-3x-9)=(1/2401)^(5x+2)

$2401^{- 3 x - 9} = {(\frac{1}{2401})}^{5 x + 2}$

Step 1

Apply the product rule to $\frac{1}{2401}$ .

$2401^{- 3 x - 9} = \frac{1^{5 x + 2}}{2401^{5 x + 2}}$

Step 2

One to any power is one.

$2401^{- 3 x - 9} = \frac{1}{2401^{5 x + 2}}$

Step 3

Move $2401^{5 x + 2}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$2401^{- 3 x - 9} = 2401^{- (5 x + 2)}$

Step 4

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

$- 3 x - 9 = - (5 x + 2)$

Step 5

Solve for $x$ .

Simplify $- (5 x + 2)$ .

Rewrite.

$- 3 x - 9 = 0 + 0 - (5 x + 2)$

Simplify by adding zeros.

$- 3 x - 9 = - (5 x + 2)$

Apply the distributive property.

$- 3 x - 9 = - (5 x) - 1 \cdot 2$

Multiply.

Multiply $5$ by $- 1$ .

$- 3 x - 9 = - 5 x - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$- 3 x - 9 = - 5 x - 2$

Move all terms containing $x$ to the left side of the equation.

Add $5 x$ to both sides of the equation.

$- 3 x - 9 + 5 x = - 2$

Add $- 3 x$ and $5 x$ .

$2 x - 9 = - 2$

Move all terms not containing $x$ to the right side of the equation.

Add $9$ to both sides of the equation.

$2 x = - 2 + 9$

Add $- 2$ and $9$ .

$2 x = 7$

Divide each term in $2 x = 7$ by $2$ and simplify.

Divide each term in $2 x = 7$ by $2$ .

$\frac{2 x}{2} = \frac{7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Divide $x$ by $1$ .

$x = \frac{7}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{7}{2}$

Decimal Form:

$x = 3.5$

Mixed Number Form:

$x = 3 \frac{1}{2}$

Solve for x 2401^(-3x-9)=(1/2401)^(5x+2)

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