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Question
math 102 exam 3 version oc summer 2006
- (8 points) simplify fully.
(a) (e^{ln 3x})
(b) (log 10^{7x})
(c) (log_2 8^{3x})
(d) (6^{log_6 2x})
- (6 points) evaluate.
(a) (log 10000)
(b) (log_{0.5} 16)
(c) (log_6 36)
- (5 points) convert to an exponential equation.
(log_4 64 = 3)
Step1: Simplify 1(a) using inverse properties
$$e^{\ln 3x} = 3x$$
Step2: Simplify 1(b) using common logarithm properties
$$\log 10^{7x} = 7x$$
Step3: Simplify 1(c) by rewriting the base
$$\log_2 8^{x} = \log_2 (2^3)^x = \log_2 2^{3x} = 3x$$
Step4: Simplify 1(d) using inverse properties
$$6^{\log_6 2x} = 2x$$
Step5: Evaluate 2(a) using powers of 10
$$\log 10000 = \log 10^4 = 4$$
Step6: Evaluate 2(b) using powers of 2
$$\log_{0.5} 16 = \log_{2^{-1}} 2^4 = -4$$
Step7: Evaluate 2(c) using fractional exponents
$$\log_{216} 36 = \log_{6^3} 6^2 = \frac{2}{3}$$
Step8: Convert 3 to exponential form
$$\log_4 64 = 3 \implies 4^3 = 64$$
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