# A few more log questions

• Mar 31st 2010, 09:34 PM
Paymemoney
A few more log questions
Hi
Need help on the following:
1)Make the independent variable the subject of the equation:
$\displaystyle u(t) = \frac{ae^t + b}{ce^t + d}$

This is what i have done, however i don't understand how the book's answer got $\displaystyle ln(\frac{du-b}{-cu+a})$

$\displaystyle u(ce^t + d) = ae^t +b$
$\displaystyle uce^t + ud - ae^t - b =0$
$\displaystyle e^t(uc - a) = -ud+b$
$\displaystyle e^t = \frac{-(ud - b)}{uc - a}$
$\displaystyle t=ln{-(ud - b)}{(uc - a)}$

2)Simply: $\displaystyle \frac{(e^x)^2}{e^{2x-1}}$
Not sure how to start this.

3)Solve for x given that: $\displaystyle ln(x) = ln(a) +n * ln(t)$
This is what i have done:

$\displaystyle ln(x) = 1 + n * ln(at)$
$\displaystyle e^{ln(x)} = e^{ln(at^{1+n}}$
$\displaystyle x=at^{1+n}$

book's answer is $\displaystyle at^n$

4)Solve for x given: $\displaystyle ln(x) = ln(a) + kt$
This is what i have done:

$\displaystyle e^{ln(x)} = e^{ln(a)} + e^{kt}$

$\displaystyle x = a + e^{kt}$

book's answer says its $\displaystyle ae^{kt}$

P.S
• Mar 31st 2010, 10:45 PM
Hello Paymemoney
Quote:

Originally Posted by Paymemoney
Hi
Need help on the following:
1)Make the independent variable the subject of the equation:
$\displaystyle u(t) = \frac{ae^t + b}{ce^t + d}$

This is what i have done, however i don't understand how the book's answer got $\displaystyle ln(\frac{du-b}{-cu+a})$

$\displaystyle u(ce^t + d) = ae^t +b$
$\displaystyle uce^t + ud - ae^t - b =0$
$\displaystyle e^t(uc - a) = -ud+b$
$\displaystyle e^t = \frac{-(ud - b)}{uc - a}$

You are correct down to here. But what's this?
Quote:

$\displaystyle t=ln{-(ud - b)}{(uc - a)}$
Surely you mean:
$\displaystyle t=\ln\left(\frac{-(ud - b)}{(uc - a)}\right)$
which is equivalent to the answer in the book. (Just multiply top-and-bottom by $\displaystyle -1$.)

Quote:

2)Simply: $\displaystyle \frac{(e^x)^2}{e^{2x-1}}$
Not sure how to start this.
Just use the rules of indices:
$\displaystyle \big(x^a\big)^b=x^{ab}$
and:
$\displaystyle \frac{x^a}{x^b}=x^{a-b}$
to write:
$\displaystyle \frac{(e^x)^2}{e^{2x-1}}=\frac{e^{2x}}{e^{2x-1}}$
$\displaystyle =e^{2x-(2x-1)}$

$\displaystyle =e$
Quote:

3)Solve for x given that: $\displaystyle ln(x) = ln(a) +n * ln(t)$
I'm not sure how you got this:
Quote:

$\displaystyle ln(x) = 1 + n * ln(at)$
On the RHS, use the rule:
$\displaystyle n\ln (y) = \ln(y^n)$
to get:
$\displaystyle \ln(x) = \ln(a) +n\ln(t)$

$\displaystyle \Rightarrow \ln(x) = \ln(a) + \ln(t^n)$
And then use:
$\displaystyle \ln(a) + \ln(b) = \ln(ab)$
to get:
$\displaystyle \Rightarrow \ln(x) = \ln(at^n)$

$\displaystyle \Rightarrow x = at^n$
Quote:

4)Solve for x given: $\displaystyle ln(x) = ln(a) + kt$
This is what i have done:

$\displaystyle e^{ln(x)} = e^{ln(a)} + e^{kt}$
No. You must combine the RHS into a single logarithm:
$\displaystyle \ln(x) = \ln(a) + kt$

$\displaystyle \Rightarrow \ln(x) = \ln(a) + kt\ln(e)$, using the fact that $\displaystyle \ln(e) = 1$.
$\displaystyle =\ln(a) +\ln(e^{kt})$

$\displaystyle =\ln(ae^{kt})$
$\displaystyle \Rightarrow x = ae{kt}$