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Inequalities – Inequality with exponential functions – Exercise 5698

Exercise

Solve the inequality:

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

Final Answer


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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