Question
Solving the equation C = Ba ^ (t/D) - K for t using logarithms with base a gives you:
A. D loga (C/B +K)
B. loga (D(C+K)/B)
C. 1/D loga (C/B) +K
D. (loga ((C+K)/B)) / D
E. D loga ((C+K)/B)
I was able to solve up to (t/D) lna = (C+K)/B
A. D loga (C/B +K)
B. loga (D(C+K)/B)
C. 1/D loga (C/B) +K
D. (loga ((C+K)/B)) / D
E. D loga ((C+K)/B)
I was able to solve up to (t/D) lna = (C+K)/B
Answers
actually, since the base is a, not e,
t/D = log_a (C+K)/B
t = D log_a ((C+K)/B)
so, choice E
t/D = log_a (C+K)/B
t = D log_a ((C+K)/B)
so, choice E
oobleck is right just did it on APEX Calculus
True Im on apex too thats correct
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