Raising both sides of equation to a common base

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The discussion centers around the correct application of exponentiation and logarithmic properties in the equation $$ln|y| = kt + C$$. The participants clarify that the correct transformation should yield $$y = Ae^{kt}$$, where $$A = e^C$$. It is emphasized that both sides of the equation must be treated as a whole when applying functions, ensuring that $$e^{kt + C}$$ is correctly expressed as $$Ae^{kt}$$. Additionally, the importance of specifying that $$A > 0$$ is highlighted, although the use of absolute values allows for $$A$$ to be any real number.

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find_the_fun
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I've been doing something very wrong for a long time.
$$
ln|y| = kt+C $$
$$e^{ln|y|}=e^{kt}+e^C $$
$$ y=e^{kt}+e^c$$
which should have been
$$
e^{ln|y|}=e^{kt+C} $$
$$ y=e^{kt} \cdot e^C = Ae^{kt}$$

Is the idea you are operating on the entire RHS and LHS side as a whole?
 
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find_the_fun said:
I've been doing something very wrong for a long time.
$$
ln|y| = kt+C $$
$$e^{ln|y|}=e^{kt}+e^C $$
$$ y=e^{kt}+e^c$$
which should have been
$$
e^{ln|y|}=e^{kt+C} $$
$$ y=e^{kt} \cdot e^C = Ae^{kt}$$

Is the idea you are operating on the entire RHS and LHS side as a whole?

Yes, that is correct. In general we have that if $x = y$ then $f(x) = f(y)$ for any function $f$ (with suitable domain). Hence the entire LHS and RHS must be passed into the function, or it doesn't work. In this case we have:

$$e^{kt + C} = e^{kt} e^C = e^C e^{kt}$$

Where the $e^C$ term can then be thought of as another constant $A = e^C$, giving:

$$e^{kt + C} = A e^{kt}$$

Intuitively, exponentiation turns addition into multiplication, and multiplication into exponentiation. Conversely, logarithms turn exponentiation into multiplication, and multiplication into addition (and addition into... addition. you can't simplify $\log(a + b)$ in general).
 
The way I look at converting from logarithmic to exponential form is to use:

$$\log_a(b)=c\implies b=a^c$$
 
Pet peeve warning! Danger Will Robinson! Danger!

Note that [math]e^C = A[/math] implies A > 0, so you need to specify that when you list your final answer.

End of Pet Peeve.

-Dan
 
topsquark said:
Pet peeve warning! Danger Will Robinson! Danger!

Note that [math]e^C = A[/math] implies A > 0, so you need to specify that when you list your final answer.

End of Pet Peeve.

-Dan

With the absolute value on the argument of the log function, I would say we need not specify a restriction on $A$...also we likely have eliminated a trivial solution $y\equiv0$ and so this can be accounted for by letting $A$ be any real number. :D

However, you do raise a good point...we should be aware of any restrictions we may impose.
 

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