A few more log questions

Hello Paymemoney

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Originally Posted by Paymemoney

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

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

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

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

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$t=ln{-(ud - b)}{(uc - a)}$

Surely you mean:

$t=\ln\left(\frac{-(ud - b)}{(uc - a)}\right)$

which is equivalent to the answer in the book. (Just multiply top-and-bottom by $-1$ .)

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

Just use the rules of indices:
$\big(x^a\big)^b=x^{ab}$
and:
$\frac{x^a}{x^b}=x^{a-b}$
to write:
$\frac{(e^x)^2}{e^{2x-1}}=\frac{e^{2x}}{e^{2x-1}}$
$=e^{2x-(2x-1)}$

$=e$

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3)Solve for x given that: $ln(x) = ln(a) +n * ln(t)$

I'm not sure how you got this:

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$ln(x) = 1 + n * ln(at)$

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

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

$\Rightarrow x = at^n$

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

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

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

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

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