Solve for x 0.5^(x+2)>9

${0.5}^{x + 2} > 9$

Step 1

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

$\ln ({0.5}^{x + 2}) > \ln (9)$

Step 2

Expand $\ln ({0.5}^{x + 2})$ by moving $x + 2$ outside the logarithm.

$(x + 2) \ln (0.5) > \ln (9)$

Step 3

Simplify the left side.

Apply the distributive property.

$x \ln (0.5) + 2 \ln (0.5) > \ln (9)$

Step 4

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

$x \ln (0.5) + 2 \ln (0.5) - \ln (9) = 0$

Step 5

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

Subtract $2 \ln (0.5)$ from both sides of the equation.

$x \ln (0.5) - \ln (9) = - 2 \ln (0.5)$

Add $\ln (9)$ to both sides of the equation.

$x \ln (0.5) = - 2 \ln (0.5) + \ln (9)$

Step 6

Divide each term in $x \ln (0.5) = - 2 \ln (0.5) + \ln (9)$ by $\ln (0.5)$ and simplify.

Divide each term in $x \ln (0.5) = - 2 \ln (0.5) + \ln (9)$ by $\ln (0.5)$ .

$\frac{x \ln (0.5)}{\ln (0.5)} = \frac{- 2 \ln (0.5)}{\ln (0.5)} + \frac{\ln (9)}{\ln (0.5)}$

Simplify the left side.

Cancel the common factor of $\ln (0.5)$ .

Cancel the common factor.

$\frac{x \ln (0.5)}{\ln (0.5)} = \frac{- 2 \ln (0.5)}{\ln (0.5)} + \frac{\ln (9)}{\ln (0.5)}$

Divide $x$ by $1$ .

$x = \frac{- 2 \ln (0.5)}{\ln (0.5)} + \frac{\ln (9)}{\ln (0.5)}$

Simplify the right side.

Cancel the common factor of $\ln (0.5)$ .

Cancel the common factor.

$x = \frac{- 2 \ln (0.5)}{\ln (0.5)} + \frac{\ln (9)}{\ln (0.5)}$

Divide $- 2$ by $1$ .

$x = - 2 + \frac{\ln (9)}{\ln (0.5)}$

Step 7

The solution consists of all of the true intervals.

$x < - 2 + \frac{\ln (9)}{\ln (0.5)}$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x < - 2 + \frac{\ln (9)}{\ln (0.5)}$

Interval Notation:

$(- \infty, - 2 + \frac{\ln (9)}{\ln (0.5)})$

Step 9

Solve for x 0.5^(x+2)>9

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