Solve for x 10^(-3x)*10^x=1/10

$10^{- 3 x} \cdot 10^{x} = \frac{1}{10}$

Step 1

Raise $10$ to the power of $1$ .

$10^{- 3 x} \cdot 10^{x} = \frac{1}{10^{1}}$

Step 2

Move $10^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$10^{- 3 x} \cdot 10^{x} = 10^{- 1}$

Step 3

Multiply $10^{- 3 x}$ by $10^{x}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$10^{- 3 x + x} = 10^{- 1}$

Add $- 3 x$ and $x$ .

$10^{- 2 x} = 10^{- 1}$

Step 4

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

$- 2 x = - 1$

Step 5

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.

Dividing two negative values results in a positive value.

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.5$