Solve the Inequality for x 0.49^x>0.7^(x+1)

${0.49}^{x} > {0.7}^{x + 1}$

Step 1

Take the log of both sides of the inequality.

$\ln ({0.49}^{x}) > \ln ({0.7}^{x + 1})$

Step 2

Expand $\ln ({0.49}^{x})$ by moving $x$ outside the logarithm.

$x \ln (0.49) > \ln ({0.7}^{x + 1})$

Step 3

Expand $\ln ({0.7}^{x + 1})$ by moving $x + 1$ outside the logarithm.

$x \ln (0.49) > (x + 1) \ln (0.7)$

Step 4

Solve the inequality for $x$ .

Simplify $(x + 1) \ln (0.7)$ .

Apply the distributive property.

$x \ln (0.49) > x \ln (0.7) + 1 \ln (0.7)$

Multiply $\ln (0.7)$ by $1$ .

$x \ln (0.49) > x \ln (0.7) + \ln (0.7)$

Subtract $x \ln (0.7)$ from both sides of the inequality.

$x \ln (0.49) - x \ln (0.7) > \ln (0.7)$

Factor $x$ out of $x \ln (0.49) - x \ln (0.7)$ .

Factor $x$ out of $x \ln (0.49)$ .

$x (\ln (0.49)) - x \ln (0.7) > \ln (0.7)$

Factor $x$ out of $- x \ln (0.7)$ .

$x (\ln (0.49)) + x (- 1 \ln (0.7)) > \ln (0.7)$

Factor $x$ out of $x (\ln (0.49)) + x (- 1 \ln (0.7))$ .

$x (\ln (0.49) - 1 \ln (0.7)) > \ln (0.7)$

Rewrite $- 1 \ln (0.7)$ as $- \ln (0.7)$ .

$x (\ln (0.49) - \ln (0.7)) > \ln (0.7)$

Divide each term in $x (\ln (0.49) - \ln (0.7)) > \ln (0.7)$ by $\ln (0.49) - \ln (0.7)$ and simplify.

Divide each term in $x (\ln (0.49) - \ln (0.7)) > \ln (0.7)$ by $\ln (0.49) - \ln (0.7)$ .

$\frac{x (\ln (0.49) - \ln (0.7))}{\ln (0.49) - \ln (0.7)} > \frac{\ln (0.7)}{\ln (0.49) - \ln (0.7)}$

Simplify the left side.

Cancel the common factor of $\ln (0.49) - \ln (0.7)$ .

Cancel the common factor.

$\frac{x (\ln (0.49) - \ln (0.7))}{\ln (0.49) - \ln (0.7)} > \frac{\ln (0.7)}{\ln (0.49) - \ln (0.7)}$

Divide $x$ by $1$ .

$x > \frac{\ln (0.7)}{\ln (0.49) - \ln (0.7)}$

Step 5

The solution consists of all of the true intervals.

$x < \frac{\ln (0.7)}{\ln (0.49) - \ln (0.7)}$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{\ln (0.7)}{\ln (0.49) - \ln (0.7)}$

Interval Notation:

$(- \infty, \frac{\ln (0.7)}{\ln (0.49) - \ln (0.7)})$

Step 7

Solve the Inequality for x 0.49^x>0.7^(x+1)

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