Log Help Please

[Corbeau_dubh]

Any light on this question would be greatly appreciated


The velocity, ?, of a vehicle during the application of brakes according to time ? is given by

?=10?−?? (that's 10e to the power of -kt)

where ? is a friction constant of the brakes. At ?=30?, the velocity is reduced to 2.5??−1. Determine ?, and state in standard form correct to two significant figures

 

[Corbeau_dubh]

Got the answer to be 0.046ms-1 (2 s.f) can anyone confirm?

 

[Mr. Bland]

Logarithms are used to solve for missing exponents. Given the following relationship:

[MATH]a^b = c[/MATH]​

The following is also true:

[MATH]log_a(c) = b[/MATH]​

We can rework the equation to solve for [MATH]k[/MATH]:

[MATH]v=10e^{-kt}[/MATH]​

​

[MATH]\frac{v}{10} = e^{-kt}[/MATH]​

​

[MATH]log_e\left(\frac{v}{10}\right) = -kt[/MATH]​

​

[MATH]k = -\frac{log_e\left(\frac{v}{10}\right)}{t}[/MATH]​

Note that [MATH]log_e[/MATH] is more commonly written as [MATH]ln[/MATH].

__________

EDIT:
Wolfram|Alpha disagrees with me, which usually means I did something wrong. I don't see it. Halp?

 

[Dr.Peterson]

[MATH]k = -\frac{log_e\left(\frac{v}{10}\right)}{t}[/MATH]
Note that [MATH]log_e[/MATH] is more commonly written as [MATH]ln[/MATH].
__________
EDIT:
Wolfram|Alpha disagrees with me, which usually means I did something wrong. I don't see it. Halp?

They don't disagree; you just didn't tell them you wanted only the real solution.

Of course, you (properly) didn't show the whole answer to the problem, which is good.