Exponential Function

**rms390** · August 23rd 2010, 03:23 AM

I have two here.

a=b(e^(-c))
solve for c

For that one, I have looked around and tried to figure it out and all i can think is that I need to use Log to solve it.

a=b/(c^2)
explain what happens to the value of a if c is doubled.

**Prove It** · August 23rd 2010, 03:37 AM

$a = b\,e^{-c}$

$\frac{a}{b} = e^{-c}$

$\ln{\left(\frac{a}{b}\right)} = -c$

$c = -\ln{\left(\frac{a}{b}\right)}$ .

**Unknown008** · August 23rd 2010, 03:43 AM

For the second one, can take the sign of proportionality.

$a = \frac{b}{c^2}$

$a\ \alpha\ \frac{1}{c^2}$

If c doubles, then;

So, if a becomes a' and c becomes c', it becomes:

$a'\ \alpha\ \frac{1}{c'^2}$

Now, we know that c' = 2c.

$a'\ \alpha\ \frac{1}{(2c)^2}$

$a'\ \alpha\ \frac{1}{4c^2}$

$4a'\ \alpha\ \frac{1}{c^2}$

So, a = 4a' and hence, a/4 = a'. You conclude that the new a' is quadrupled when compared to the previous a

(here, c' means the new value of c, not the derivative of c.)

[Note, until I find the sign for the proportionality, I'll use alpha instead]

**rms390** · August 23rd 2010, 04:22 AM

thank you very much

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