Solve for x 7^x> cube root of 7

$7^{x} > \sqrt[3]{7}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{7}$ as $7^{\frac{1}{3}}$ .

$7^{x} > 7^{\frac{1}{3}}$

Step 2

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$x > \frac{1}{3}$

Step 3

The result can be shown in multiple forms.

Inequality Form:

$x > \frac{1}{3}$

Interval Notation:

$(\frac{1}{3}, \infty)$

Step 4