I am fairly well acquainted with logs, yet when i came across this, I couldn't understand the answer.
4. If \(\displaystyle P\, =\, Ke^{-xt},\) then \(\displaystyle x\) equals:
(A) \(\displaystyle \dfrac{\log\, K}{t\, log\,e \log\, P}\)
(B) \(\displaystyle \dfrac{P}{Ke^t}\)
(C) \(\displaystyle \dfrac{Pe^t}{K}\)
(D) \(\displaystyle \dfrac{\log\, K\, -\, \log\, P}{t\, \log\, e}\)
(E) none of these
The correct answer is (D), with this solution provided:
\(\displaystyle \dfrac{P}{K}\, =\, e^{-xt}\) or \(\displaystyle -tx\, =\, \log_e\, \dfrac{P}{K}\)
\(\displaystyle x\, =\, -\dfrac{1}{t}\left(\log_e\, P\, -\, \log_e\, K\right)\, =\, \dfrac{\log_e\, K\, -\, \log_e\, P}{t}\)
\(\displaystyle x\, =\, \dfrac{\log\, K\, -\, \log\, P}{t\, \log\, e}\)
Can you solve the problem in even more detail than this and tell me what rules/formulas are used?
4. If \(\displaystyle P\, =\, Ke^{-xt},\) then \(\displaystyle x\) equals:
(A) \(\displaystyle \dfrac{\log\, K}{t\, log\,e \log\, P}\)
(B) \(\displaystyle \dfrac{P}{Ke^t}\)
(C) \(\displaystyle \dfrac{Pe^t}{K}\)
(D) \(\displaystyle \dfrac{\log\, K\, -\, \log\, P}{t\, \log\, e}\)
(E) none of these
The correct answer is (D), with this solution provided:
\(\displaystyle \dfrac{P}{K}\, =\, e^{-xt}\) or \(\displaystyle -tx\, =\, \log_e\, \dfrac{P}{K}\)
\(\displaystyle x\, =\, -\dfrac{1}{t}\left(\log_e\, P\, -\, \log_e\, K\right)\, =\, \dfrac{\log_e\, K\, -\, \log_e\, P}{t}\)
\(\displaystyle x\, =\, \dfrac{\log\, K\, -\, \log\, P}{t\, \log\, e}\)
Can you solve the problem in even more detail than this and tell me what rules/formulas are used?
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