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Solve This Maths Problem?
- Thread starter Shaun.P
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No. The question appears in a book before the concept of the natural log is introduced and I was wondering if it was misplaced. I therefore wondered if I was missing something.
I'm not a physicist so I've never solved this particular equation. I do regularly do pharmaceutical calculations (pharmacokinetics) which often involve exponentials, so this is similar and very simple. Usually I deal with concentration of a drug in blood over time. I can hopefully give you enough info get you moving in the tight direction.
The trial and error method could solve anything if you have enough time. Impress your teacher and just modify the equation (shown below), plug in the values, and use the LN button on your calculator.
Ln[Q(t)] = Ln(Q0) - (T/RC)
Ln is the inverse of e^x
Therefore multiplying Ln by e^x will give you X
The concept behind natural logs is a little complicated but I'm sure you can find a great explanation on the internet. I don't know what these specific variables stand for (I'm not a physicist/electronics guru), I assume you do, so you should be all set in solving the equation.
I hope that helps?
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Edit: Regarding the shape of the graph, the application of the Ln will take an exponential equation and make it linear. The equation isn't changed because you're doing the same thing to both sides, but your results will appear differently. Later in some course you may find this to be handy in analysis because it ultimately allows us to take exponential data and analyze easily in a linear fashion (prediction of values, linear regression, etc). Just know the data derived will have to be reconverted using it's exponential (e^, the inverse of Ln) to have a proper (actual in reality) value.
Off the top of my head the slope of the graph will be determined on whether or not the rate constant is + or - in the equation. If the exponent value turns out to be a negative value, then the slope will be negative. If the exponent ends up being positive, the slope will be positive.
I'm sure there are some true mathematicians who can explain this better. Again, this is a very basic, general explanation that like anything in math can get more and more complicated. Someone with knowledge of the specific science you're working in will better able to describe this in a conceptual sense knowing the variables and understanding how electricity works. I'm not that guy.
I have attached below a graph comparing 2^x and Ln(2^x). You'll see the Ln(2^x) has become linear. As you can see with both the graph and the physics equation, once you apply Ln, the exponential is removed.
Also attached some important rules of dealing with Natural Logs.
Hope this helped?
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Wait, didn't you just use Ln? While the OP asked "is it possible to solve this without the use of Ln?"
Wait, didn't you just use Ln? While the OP asked "is it possible to solve this without the use of Ln?"
Yes
Wait, didn't you just use Ln? While the OP asked "is it possible to solve this without the use of Ln?"
Then he said
So I wasn't sure if he didn't know how to use natural logs since "the concept of the natural log is not introduced". I took that as not knowing how to properly use natural logs which probably isn't uncommon. If my assumption was correct, sorry for helping the person figure out his homework on his own like an intelligent, self starting student rather than be a drone and wait to be told how to do it tomorrow. He could at least make some inferences on the shape of the graph.
If he does know how to use Ln, he shouldn't have a problem solving the problem and I'd wonder why he'd ask the question.
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Posts #1-#3 suggested to me that the goal was to ascertain whether there was a straightforward way to solve the problem without a knowledge of natural logs.
That was not blatantly evident to me as it was to you.
Alright, I see he's a teacher now, not a student. Oooookayyyyy then...
Nevermind
That was not blatantly evident to me as it was to you.
Alright, I see he's a teacher now, not a student. Oooookayyyyy then...
Nevermind
Hey, it was pretty nice of you to type that all up...
Hey, it was pretty nice of you to type that all up...
I try. I'm a fan of people learning by doing and asking questions along the way. I admit I sometimes read too far into things. This is because growing up my father rather than saying "can you mow the lawn" would say "it looks like the grass is getting deep." People can phrase questions in round about ways.
Ok, now bearing in mind we're not literally trying to solve the problem:
If the student understands exponential equations then yes. If not, then no.
If they understand e= 2.7182... or at the very least is a positive number:
1) I assume they know if the variables they are using are always + or - or absolute values or whatever... therefore, if all variables are + (or absolute values) then they should know this is a exponential curve with a negative slope based on the negative exponential. By sketch that could mean a rough diagram or actually plugging in numbers. Either way that should be solvable.
2) I don't believe there would be a way to solve the problems without using Ln, though I don't fully understand all the variables in play. If the student doesn't understand Ln, they probably shouldn't have some weird backwards way of solving this. Otherwise they deserve to be in a higher level math class.
I'm not sure how things are taught, I don't remember honestly, but if they don't understand Ln will they understand e^? I feel like those things would be taught in conjunction, unless the OP has already specifically taught them only how to use e^.
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There is a solution without ln that uses another logarithm. Does this count, i.e. was the (general) logarithmic function part of the book yet?
The given problem states an equation using e. I thought e and ln go hand-in-hand by definition? How can a problem using e pre-date the natural log?
The given problem states an equation using e. I thought e and ln go hand-in-hand by definition? How can a problem using e pre-date the natural log?
The problem didn't state that the constant k has to be a real number. It could be a term consisting of other mathematical expressions.
It's not entirely clear to me where imaginary numbers would pop up in this context. Although perhaps you're unclear about what a real number is?
Thanks for the input folks.
The answer is 0.7RC and is found by taking the natural log.
This exercise is from a book where the concept of e is first introduced and in most of the questions involved substituting in numbers in order to find what the equations were equal to. This question obviously asks for a power but the rest of the exercise didn't do this.
A colleague and I feel the question is misplaced within the textbook but just wanted to check with some others first.
The answer is 0.7RC and is found by taking the natural log.
This exercise is from a book where the concept of e is first introduced and in most of the questions involved substituting in numbers in order to find what the equations were equal to. This question obviously asks for a power but the rest of the exercise didn't do this.
A colleague and I feel the question is misplaced within the textbook but just wanted to check with some others first.
Hopefully it's a very old book. I haven't heard the name "condenser" used for a capacitor in over 50 years.
No, you misunderstood me. I meant the solution needn't be like 0.7 but could be e.g. log 2/log e, which has of course a real solution, but isn't a real number in itself.
Definitely a real number.
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Yeah, if the students haven't studied logarithms yet it certainly seems out of place.
In other words: You can compute the 0.7 (heavily rounded btw) without using ln by using another logarithmic base. I just used the term above
(log 2 / log e = 0.7) to make it clear that ln isn't needed.
I unfortunately had the problem explaining what I meant in English.
(log 2 / log e = 0.7) to make it clear that ln isn't needed.
I unfortunately had the problem explaining what I meant in English.
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