Question

in exercises 47–54, solve the inequality. (see examples 5 and 6.)

1. ( 9^{x} > 54 )
2. ( 4^{x} leq 36 )
3. ( ln x geq 3 )
4. ( log_{4} x < 4 )
5. ( 3^{x - 5} < 8 )
6. ( e^{3x - 4} > 11 )
7. ( -3log_{2} x - 6 leq 9 )
8. ( -4log_{3} x - 5 geq 3 )

Explanation:

Problem 47: $9^x > 54$

Step1: Take log base 9 of both sides

$\log_9(9^x) > \log_9(54)$

Step2: Simplify using $\log_b(b^x)=x$

$x > \log_9(54)$

Step3: Rewrite in base 3 (or natural log)

$x > \frac{\ln(54)}{\ln(9)} = \frac{\ln(6 \cdot 9)}{\ln(9)} = \frac{\ln6 + \ln9}{\ln9} = 1 + \frac{\ln6}{2\ln3}$

Step1: Take log base 4 of both sides

$\log_4(4^x) \leq \log_4(36)$

Step2: Simplify using $\log_b(b^x)=x$

$x \leq \log_4(36)$

Step3: Rewrite in base 2 (or natural log)

$x \leq \frac{\ln(36)}{\ln(4)} = \frac{2\ln6}{2\ln2} = \log_2 6$

Step1: Exponentiate both sides with base $e$

$e^{\ln x} \geq e^3$

Step2: Simplify using $e^{\ln x}=x$

$x \geq e^3$

Answer:

$x > \log_9 54$ or $x > 1 + \frac{\ln 6}{2\ln 3}$

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Problem 48: $4^x \leq 36$