Solving for time in an exponential equation

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Homework Statement



In my book, there is a formula that gives the amount (in grams) of Radium in a jar after t years (100 grams were initially stored):

R = 100⋅e-0.00043⋅t

The book asks me to sketch the graph of the equation. I decided to find a point where the time elapsed equals the remaining radium to use as a point for the sketch. This is were I found a problem.

Homework Equations



The properties of logarithms are relevant.

The Attempt at a Solution


[/B]
The equation is:

R = 100⋅e-0.00043⋅t

I am looking for the point in time where t = R. I substitute:

t = 100⋅e-0.00043⋅t

Then I divide by e-0.00043⋅t on both sides:

t/e-0.00043⋅t = 100

Take the natural log of both sides:

ln (t/e-0.00043⋅t) = ln (100)

Quotient property:

ln(t) - ln (e-0.00043⋅t) = ln(100)

ln (t) + 0.00043⋅t = ln (100)

Here is where I hit a block. I have t inside the logarithm. To get rid of the logarithm, I have raise e to both sides of the equation. If I do that, I make the t outside the logarithm an exponent while the other one isn't. I can't seem to find a way to bring the both "t"s together so that I can solve. Please help. Thanks. If I'm not clear somewhere please tell me.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



In my book, there is a formula that gives the amount (in grams) of Radium in a jar after t years (100 grams were initially stored):

R = 100⋅e-0.00043⋅t

The book asks me to sketch the graph of the equation. I decided to find a point where the time elapsed equals the remaining radium to use as a point for the sketch.
Why would you do that? Time is measured in years and radium in grams. It's like asking when I am as old as my weight. For drawing the graph it would make much more sense to pick a time when the radium is half of what it started or something similar. Not to mention it would give you an equation you can solve directly with logarithms.
 
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  • #3
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The reason I am finding the point at which they are equal is that I also am trying to find a method to solve equations in the form x = abx. While it isn't the most practical way to solve my problem, I would think it is useful to know when solving systems of equations with linear and exponential equations (Ex. y = ax and y = bx). Rather than having to solve by making tables, one could solve algebraically.
 
  • #4
berkeman
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The reason I am finding the point at which they are equal is that I also am trying to find a method to solve equations in the form x = abx. While it isn't the most practical way to solve my problem, I would think it is useful to know when solving systems of equations with linear and exponential equations (Ex. y = ax and y = bx). Rather than having to solve by making tables, one could solve algebraically.
If you check the Profile page for @LCKurtz you will see that he has a PhD in math (click on his avatar). Maybe you should consider taking his advice on the best way to approach this problem... :smile:
 
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  • #5
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Thanks! I saw that and I found his advice very helpful, I didn't mean to come off as arrogant. While my big question was answered, I am just curious about a way to solve the equation as it is (as impractical as it may be :) )
 
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  • #6
berkeman
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Thanks! I saw that and I found his advice very helpful, I didn't mean to come off as arrogant. While my big question was answered, I am just curious about a way to solve the equation as it is (as impractical as it may be :) )
It didn't seem arrogant to me at all, in fact you seemed to be asking a good question about a "more general" equation...
R = 100⋅e-0.00043⋅t
I also am trying to find a method to solve equations in the form x = ab^x.
Those seemed like fairly different problems to me. Solving exponential equations involving e^x type of terms and ln(x) are much more common in my experience than the unusual equation you posted. Have you seen the form that you posted somewhere?
 
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Yes they are different problems. Should I make a different form?
Have you seen the form that you posted somewhere?
Which form are you talking about?
 
  • #8
berkeman
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Yes they are different problems. Should I make a different form?
A different thread? Maybe, depending on your answer to the next question pair... :smile:
Which form are you talking about?
x = ab^x
That one. What class of problems is that from? As an EE, my math background is limited to undergraduate math, and dealing with the math that EEs deal with.
 
  • #9
LCKurtz
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If you want do know how to solve functions like ##x=ab^x##, look up the Lambert W function. It's basic definition is if ##y = xe^x## then ##x = W(y)##, the Lambert W function. It is not an elementary function, but you can solve problems like ##x=ab^x## in terms of it. I will show you how in a few minutes.
 
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  • #10
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Thank you. I'll look at that.
 
  • #11
LCKurtz
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##\begin{array}{rcl}
x & = & ab^x\\
xb^{-x} & = &a \\
xe^{-x\ln b} & = & a\\
x(-\ln b)e^{-x\ln b} &=& -a\ln b\\
-x\ln b &=& W(-a\ln b)\\
x & = & -\frac{W(-a\ln b)}{\ln b}
\end{array}
##
 
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  • #12
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Thank you. This helps a lot. Before I completely understand I will need to learn more but I am glad to know what method to learn and use.
 
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  • #13
Ray Vickson
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The reason I am finding the point at which they are equal is that I also am trying to find a method to solve equations in the form x = abx. While it isn't the most practical way to solve my problem, I would think it is useful to know when solving systems of equations with linear and exponential equations (Ex. y = ax and y = bx). Rather than having to solve by making tables, one could solve algebraically.
The equation is IMPOSSIBLE to solve algebraically; by that I mean that people have proved that no finite algebraic formula can possibly exist that will give you the solution to ##x = a b^x## for general values of ##a>0## and ##b>0##. Note: it is not just that nobody has been smart enough to solve the problem algebraically; it is that it has been proven to be impossible. It is absolutely necessary to use numerical methods on such problems.

There are formulas for the solution, but they involve so-called "non-elementary" functions, so in a way it just expresses the solution of one impossible equation in terms of another. Nevertheless, such forms are very useful because modern computer algebra packages do have efficient and accurate ways of doing the required calculations, and lots of properties are known for the non-elementary function involved. (So, in modern computer algebra systems, using such a function is just as easy and straightforward as using a trigonometric function or a square root!)

For the record: Maple gets the solution of the equation ##x = a b^x## as
$$x = -\frac{1}{\ln (b)} \text{LambertW}(-a \, \ln(b)),$$
where ##\ln(\cdot)## is the natural logarithm and LambertW is the so-called Lambert W-function.

Note added in edit: for some reason, all the responses from #9 to #12 did not appear on my screen until after I pressed the "enter" key to post my message! That makes my message redundant, but I'll leave it in the thread nevertheless.
 
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