MHB Raising both sides of equation to a common base

  • Thread starter Thread starter find_the_fun
  • Start date Start date
  • Tags Tags
    Base
Click For Summary
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.
find_the_fun
Messages
147
Reaction score
0
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?
 
Mathematics news on Phys.org
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.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
2K
Replies
8
Views
4K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K