Solve for x 10^(2x)=0.1

$10^{2 x} = 0.1$

Take the base $10$ logarithm of both sides of the equation to remove the variable from the exponent.

$\log (10^{2 x}) = \log (0.1)$

Expand the left side.

Expand $\log (10^{2 x})$ by moving $2 x$ outside the logarithm.

$2 x \log (10) = \log (0.1)$

Logarithm base $10$ of $10$ is $1$ .

$2 x \cdot 1 = \log (0.1)$

Multiply $2$ by $1$ .

$2 x = \log (0.1)$

Simplify the right side.

Log base $10$ of $0.1$ is approximately $- 1$ .

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Cancel the common factor of $- 1$ and $2$ .

Rewrite $- 1$ as $1 (- 1)$ .

$x = \frac{1 (- 1)}{2}$

Cancel the common factors.

Rewrite $2$ as $1 (2)$ .

$x = \frac{1 \cdot - 1}{1 \cdot 2}$

Cancel the common factor.

$x = \frac{1 \cdot - 1}{1 \cdot 2}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 0.5$