Solve for x x^3+36^(2x)=1/6+x^3

$x^{3} + 36^{2 x} = \frac{1}{6} + x^{3}$

Step 1

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

Subtract $x^{3}$ from both sides of the equation.

$x^{3} + 36^{2 x} - x^{3} = \frac{1}{6}$

Combine the opposite terms in $x^{3} + 36^{2 x} - x^{3}$ .

Subtract $x^{3}$ from $x^{3}$ .

$0 + 36^{2 x} = \frac{1}{6}$

Add $0$ and $36^{2 x}$ .

$36^{2 x} = \frac{1}{6}$

Step 2

Raise $6$ to the power of $1$ .

$36^{2 x} = \frac{1}{6^{1}}$

Step 3

Move $6^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$36^{2 x} = 6^{- 1}$

Step 4

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

$6^{2 (2 x)} = 6^{- 1}$

Step 5

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

$2 (2 x) = - 1$

Step 6

Solve for $x$ .

Multiply $2$ by $2$ .

$4 \cdot x = - 1$

Divide each term in $(4 \cdot x) = - 1$ by $4$ and simplify.

Divide each term in $(4 \cdot x) = - 1$ by $4$ .

$\frac{(4 \cdot x)}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 \cdot x}{4} = \frac{- 1}{4}$

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{4}$

Decimal Form:

$x = - 0.25$