Solve for x 7^(3x)>5

$7^{3 x} > 5$

Step 1

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

$\ln (7^{3 x}) > \ln (5)$

Step 2

Expand $\ln (7^{3 x})$ by moving $3 x$ outside the logarithm.

$3 x \ln (7) > \ln (5)$

Step 3

Divide each term in $3 x \ln (7) > \ln (5)$ by $3 \ln (7)$ and simplify.

Divide each term in $3 x \ln (7) > \ln (5)$ by $3 \ln (7)$ .

$\frac{3 x \ln (7)}{3 \ln (7)} > \frac{\ln (5)}{3 \ln (7)}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x \ln (7)}{3 \ln (7)} > \frac{\ln (5)}{3 \ln (7)}$

Rewrite the expression.

$\frac{x \ln (7)}{\ln (7)} > \frac{\ln (5)}{3 \ln (7)}$

Cancel the common factor of $\ln (7)$ .

Cancel the common factor.

$\frac{x \ln (7)}{\ln (7)} > \frac{\ln (5)}{3 \ln (7)}$

Divide $x$ by $1$ .

$x > \frac{\ln (5)}{3 \ln (7)}$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$x > \frac{\ln (5)}{3 \ln (7)}$

Interval Notation:

$(\frac{\ln (5)}{3 \ln (7)}, \infty)$

Step 5