Question

3. Solve for x a) $e^{(x+4)} = \frac{1}{e^{2x}}$ d) $3\log_2(x-1) = 5 - \log_24$

          3. Solve for x
a) $e^{(x+4)} = \frac{1}{e^{2x}}$
d) $3\log_2(x-1) = 5 - \log_24$

3. Solve for x
a) e^(x+4) = (1)/(e^2x)
d) 3log2(x-1) = 5 - log24

Added by Lisa G.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Tim Thornhill

Step 1

d) To solve the equation 3log(x - 1) = 5log24, we can start by dividing both sides of the equation by 3: 3log(x - 1) / 3 = 5log24 / 3 Simplifying, we get: log(x - 1) = (5/3)log24 Show more…

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a) e^(4) = 4e^4 d) 3log(x - 1) = 5log24

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