Exercise
Solve the inequality:
{(\frac{3}{4})}^{2x+3}<{(\frac{27}{64})}^{x+1}
Final Answer
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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