Solve for x 5(10^(x+2))=25

$5 (10^{x + 2}) = 25$

Step 1

Divide each term in $(5 \cdot 10^{x + 2}) = 25$ by $5$ and simplify.

Divide each term in $(5 \cdot 10^{x + 2}) = 25$ by $5$ .

$\frac{(5 \cdot 10^{x + 2})}{5} = \frac{25}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 \cdot 10^{x + 2}}{5} = \frac{25}{5}$

Divide $10^{x + 2}$ by $1$ .

$10^{x + 2} = \frac{25}{5}$

Simplify the right side.

Divide $25$ by $5$ .

$10^{x + 2} = 5$

Step 2

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

$\log (10^{x + 2}) = \log (5)$

Step 3

Expand the left side.

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

$(x + 2) \log (10) = \log (5)$

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

$(x + 2) \cdot 1 = \log (5)$

Multiply $x + 2$ by $1$ .

$x + 2 = \log (5)$

Step 4

Subtract $2$ from both sides of the equation.

$x = \log (5) - 2$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \log (5) - 2$

Decimal Form:

$x = - 1.30102999 \dots$