MHB Raising both sides of equation to a common base

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The discussion centers on the correct application of exponentiation and logarithms in the equation ln|y| = kt + C. It highlights the mistake of treating the right-hand side (RHS) and left-hand side (LHS) separately instead of as a whole, emphasizing that exponentiating both sides must be done uniformly. The correct transformation leads to y = A e^(kt), where A = e^C, and it's noted that A must be positive unless absolute values are considered. The conversation also touches on the implications of these transformations, including the need to specify restrictions on A based on the context of the logarithm. Overall, the importance of understanding the properties of logarithms and exponentials in solving equations is underscored.
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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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