Raising both sides of equation to a common base

[find_the_fun]

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

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

 

[Nono713]

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

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).

 

[MarkFL]

The way I look at converting from logarithmic to exponential form is to use:

\(\displaystyle \log_a(b)=c\implies b=a^c\)

 

[topsquark]

Pet peeve warning! Danger Will Robinson! Danger!

Note that \(\displaystyle e^C = A\) implies A > 0, so you need to specify that when you list your final answer.

End of Pet Peeve.

-Dan

 

[MarkFL]

Pet peeve warning! Danger Will Robinson! Danger!

Note that \(\displaystyle e^C = A\) 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.