Solve for x 105^(x+2)>1000

$105^{x + 2} > 1000$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (105^{x + 2}) > \ln (1000)$

Expand $\ln (105^{x + 2})$ by moving $x + 2$ outside the logarithm.

$(x + 2) \ln (105) > \ln (1000)$

Simplify the left side.

Apply the distributive property.

$x \ln (105) + 2 \ln (105) > \ln (1000)$

Move all the terms containing a logarithm to the left side of the equation.

$x \ln (105) + 2 \ln (105) - \ln (1000) = 0$

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

Subtract $2 \ln (105)$ from both sides of the equation.

$x \ln (105) - \ln (1000) = - 2 \ln (105)$

Add $\ln (1000)$ to both sides of the equation.

$x \ln (105) = - 2 \ln (105) + \ln (1000)$

Divide each term in $x \ln (105) = - 2 \ln (105) + \ln (1000)$ by $\ln (105)$ and simplify.

Divide each term in $x \ln (105) = - 2 \ln (105) + \ln (1000)$ by $\ln (105)$ .

$\frac{x \ln (105)}{\ln (105)} = \frac{- 2 \ln (105)}{\ln (105)} + \frac{\ln (1000)}{\ln (105)}$

Simplify the left side.

Cancel the common factor of $\ln (105)$ .

Cancel the common factor.

$\frac{x \ln (105)}{\ln (105)} = \frac{- 2 \ln (105)}{\ln (105)} + \frac{\ln (1000)}{\ln (105)}$

Divide $x$ by $1$ .

$x = \frac{- 2 \ln (105)}{\ln (105)} + \frac{\ln (1000)}{\ln (105)}$

Simplify the right side.

Cancel the common factor of $\ln (105)$ .

Cancel the common factor.

$x = \frac{- 2 \ln (105)}{\ln (105)} + \frac{\ln (1000)}{\ln (105)}$

Divide $- 2$ by $1$ .

$x = - 2 + \frac{\ln (1000)}{\ln (105)}$

The solution consists of all of the true intervals.

$x > - 2 + \frac{\ln (1000)}{\ln (105)}$

The result can be shown in multiple forms.

Inequality Form:

$x > - 2 + \frac{\ln (1000)}{\ln (105)}$

Interval Notation:

$(- 2 + \frac{\ln (1000)}{\ln (105)}, \infty)$