# How do you solve the exponential inequality 27^(4x-1)>=9^(3x+8)?

Mar 26, 2017

$x \ge \frac{9.5}{3}$

#### Explanation:

${27}^{4 x - 1} \ge {9}^{3 x + 8}$

law of indices:
$\left({a}^{m} \cdot {a}^{n} = {a}^{m + n}\right)$

$27 = {3}^{3}$
$9 = {3}^{2}$

$\therefore 27 = {9}^{\frac{3}{2}}$

${27}^{4 x - 1} = {9}^{1.5 \left(4 x - 1\right)}$

$= {9}^{6 x - 1.5}$

${9}^{6 x - 1.5} \ge {9}^{3 x + 8}$

$6 x - 1.5 \ge 3 x + 8$

add $1.5$:

$6 x \ge x + 9.5$

subtract $3 x$:

$3 x \ge 9.5$

$x \ge \frac{9.5}{3}$

Aug 10, 2018

$x = \frac{19}{6}$

#### Explanation:

It would be very nice if we had the same base. Recall that

$27 = {3}^{3}$ and $9 = {3}^{2}$

With this in mind, we can rewrite our inequality as

3^(3(4x-1))=3^(2(3x+8)

We can further simplify the exponents to get

${3}^{12 x - 3} = {3}^{6 x + 16}$

Now that our bases are the same, we can equate the exponents:

$12 x - 3 = 6 x + 16$

$6 x = 19$

$x = \frac{19}{6}$

Hope this helps!