# How do you solve the exponential inequality 4^x<=4^(3x-1)?

Jun 26, 2017

See explanation.

#### Explanation:

If the bases of both sides of the exponential equation (inequality) are equal, you can change the equation (inequality) to the equation (inequality) of the exponents.

When leaving out bases you have to be careful about the sign of rhe inequality.

• if the base is between $0$ and $1$ you change the sign of the inequality to opposite $>$ to $<$ and vice versa.

• if the base is greater than $1$ the sign stays unchanged.

In the given example the base is $4$, so the sign stays unchanged.

changes to:

## $x \le 3 x - 1$ 

$x - 3 x \le - 1$

$- 2 x \le - 1$

$x \ge \frac{1}{2}$

Jun 26, 2017

$x \ge \frac{1}{2}$

#### Explanation:

Your base on both sides are the same, so you have completely nothing to worry about there.

The only problem now is the powers:
$x = \left(3 x - 1\right)$

You now have to get all x values on one side.
$\left(x + 1\right) = 3 x$

Take away $x$ from both sides:
$1 = 2 x$

Now divide both sides by 2 to find x:
$\frac{1}{2} = x$

Proof:
${4}^{\frac{1}{2}}$ is just that.
${4}^{3 \left(\frac{1}{2}\right) - 1} = {4}^{\frac{3}{2} - 1} = {4}^{\frac{3}{2} - \frac{2}{2}} = {4}^{\frac{1}{2}}$

${4}^{\frac{1}{4}} = \sqrt[4]{4} \approx 1.41$
${4}^{3 \left(\frac{1}{4}\right) - 1} \approx 0.707$

${4}^{1} = 4$
${4}^{3 \left(1\right) - 1} = {4}^{2} = 16$

Therefore, $x \ge \left(\frac{1}{2}\right)$.