# Inequalities – Inequality with exponential functions – Exercise 5698

Exercise

Solve the inequality:

$${(\frac{3}{4})}^{2x+3}<{(\frac{27}{64})}^{x+1}$$

$$x< 0$$

Solution

$${(\frac{3}{4})}^{2x+3}<{(\frac{27}{64})}^{x+1}$$

Compare the bases:

$${(\frac{3}{4})}^{2x+3}<{(\frac{3^3}{4^3})}^{x+1}$$

Using power rules we get:

$${(\frac{3}{4})}^{2x+3}<{(\frac{3}{4})}^{3(x+1)}$$

$${(\frac{3}{4})}^{2x+3}<{(\frac{3}{4})}^{3x+3}$$

We have the same base in both sides. Because it is smaller than one, we get

$$3x+3<2x+3$$

We solve the inequality:

$$3x-2x<3-3$$

$$x<0$$

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