Logarithm problem

[Max Bacon]

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?

 

[Deleted member 4993]

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?

Exactly which line do you not understand?

I'll start you off...

From the first line (the given problem) - you divide both sides by 'K' - to get the second line $\displaystyle \dfrac{P}{K} \ = \ e^{-xt}$

for the next line you apply "log_e" to both sides. What does that operation look like?

 

[Max Bacon]

Left it, then came back later to figure it out. I had failed to solve it because i was writing log_e^-xt instead of log_ee^-xt ,(I wrote the exponent as the number) which had left me puzzled. Maybe I should have picked a different base, something simple like 10, 1000 or just log to avoid confusion.

Also, I'll remember to write out the problems in the future, instead of showing them with a pic.

Anyway, Thanks for the help.

 

[Deleted member 4993]

Left it, then came back later to figure it out. I had failed to solve it because i was writing log_e^-xt instead of log_ee^-xt ,(I wrote the exponent as the number) which had left me puzzled. Maybe I should have picked a different base, something simple like 10, 1000 or just log to avoid confusion.

Also, I'll remember to write out the problems in the future, instead of showing them with a pic.

Anyway, Thanks for the help.

I write "log_e" as "ln" and "log₁₀" as "log"

In other words, for base e and 10 - I do not write the base.

That may help to avoid confusion.

 

[HallsofIvy]

I would do that myself but you should be aware that many advanced books use "log(x)", without any base shown, to mean the natural logarithm on the grounds that there is no reason to use the common logarithm which is based on a particular numeration system.

 

[soroban]

Hello, Max Bacon!

4. Solve for $\displaystyle x\!:\;P\: =\: Ke^{-xt},$

$\displaystyle (A)\; \dfrac{\log\, K}{t\, log\,e \log\, P} \quad (B)\; \dfrac{P}{Ke^t} \quad (C)\; \dfrac{Pe^t}{K}$

$\displaystyle (D)\;\dfrac{\log\, K\, -\, \log\, P}{t\, \log\, e} \quad (E)\;\text{none of these}$

We are given: .$\displaystyle P \;=\;Ke^{-xt}$

Multiply by $\displaystyle e^{xt}\!:\;Pe^{xt} \;=\;K$

Divide by $\displaystyle P\!:\;e^{xt} \;=\;\frac{K}{P}$

Take logs: .$\displaystyle \log\left(e^{xt}\right) \;=\;\log\left(\frac{K}{P}\right)$

We have: .$\displaystyle xt\log e\;=\;\log(K) - \log(P)$ .**

Divide by $\displaystyle t\log e\!:\;x \;=\;\dfrac{\log(K) - \log (P)}{t\log e}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Two log rules were used:

. . $\displaystyle \log(x^n) \;=\;n\log x$

. . $\displaystyle \log\left(\frac{x}{y}\right) \;=\;\log x - \log y$

 

[HallsofIvy]

Equivalently, from $\displaystyle P= Ke^{-xt}$, divide both sides by K: $\displaystyle \frac{P}{K}= e^{-xt}$. Take the logarithm of both sides: $\displaystyle log(P/K)= log(P)- log(K)= -xt$. Finally, divide both sides by -t: $\displaystyle x= \frac{log(P)- log(K)}{-t}= \frac{log(K)- log(P)}{t}$. That "log(e)" in the denominator of "D" is a little peculiar. If "log" is intended to be "logarithm base e" or "natural logarithm" that would be 1. But "D" is correct for a logarithm to any base.

 

[Quaid]

why show "log(e)" in ANY expression, since it equals 1

log(e) does not equal 1, if log denotes common logarithms.

Otherwise, one answer could be: "For teaching purposes".

 

[Quaid]

That "log(e)" in the denominator of "D" is a little peculiar.

Agree!

Maybe the question and the multiple-choices were all machine-generated. If so, the programmer may have lacked insight regarding how built-in functions would handle such situations.