Lexi V.

asked • 04/08/22

Pre Calc Question

Convert the exponential equation 𝑄=0.35𝑒^0.7𝑡

to the form 𝑄=𝑎𝑏𝑡 and enter the values of a and b below.

a=

b=

Dayv O.

hi Lexi, did you find from your teacher what answer was deemed correct? thx.
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04/08/22

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Dayv O. answered • 04/08/22

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e.7=2 approximately

Q=.35*2t just about

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Donald W.

Just my own two cents, but I don't think we should be encouraging students to approximate irrational numbers with integers, especially in a pre-calculus class. Before you know it, we'll have people wanting us to use 3 instead of pi everywhere. 😂 This is especially true for an exponential equation, where the margin of error will grow, well, exponentially.
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04/08/22

Dayv O.

2 to an exponent variable is doubling, If someone was trying to convert an e base equation to a doubling equation exactly, the the exponent of e would need to b= .69314718056...My two cents is the problem is pointing out the e to .7t is very much close to doubling.
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04/08/22

Donald W.

Possibly. Maybe it's just a poor choice of question to give to a student. 🤷‍♂️
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04/08/22

Donald W.

FWIW, the error rate grows pretty quickly. For a t value of 20, 2^t is 1048576 while (e^0.7)^t is ~1202604. That's a difference of ~154028, which is almost 13%. That's why it's not a good idea to use approximations in an exponential expression. The error rate grows really quickly.
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04/08/22

Dayv O.

agree
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04/08/22

Donald W. answered • 04/08/22

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Not entirely sure what this question is asking for. Does it not like the 0.7 in the exponent and wants an equation of this form?


Q = a * bt


If so, we can rewrite our original Q equation to get there:


Q = 0.35 * e0.7*t

Q = 0.35 * (e0.7)t


So a = 0.35 and b = e0.7. e0.7 is irrational, so it can't be simplified anymore without losing precision.

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Dayv O.

but e^.7=2.013752700747...,,,,,,,,,the problem wants an answer I think, not that the answer is there is no answer. Q=.35*2.013752700747^t with a high degree of accuracy.
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04/08/22

Donald W.

Yeah, I'd throw this question back to the teacher and ask them what they're really asking for and what the point is.
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04/08/22

Donald W.

FWIW, I think b = e^0.7 is a valid answer. It's a real number.
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04/08/22

Dayv O.

question reads"Convert the exponential equation 𝑄=0.35𝑒^0.7𝑡 to the form 𝑄=𝑎𝑏𝑡 and enter the values of a and b below." and a valid answer is there is no way to convert? The question wants the variable t not multiplied by any constant other than 1.
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04/08/22

Donald W.

If a is 0.35 and b is e^0.7, then t just has a coefficient of 1. So this is a valid answer.
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04/08/22

Dayv O.

nothing is converted other than parentheses added. To fix any accuracy problem, e^.7 can be 2.01, or 2.013, or 2.0137,....
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04/08/22

Donald W.

On the contrary, we are using the rules of exponents to rewrite e^(0.7*t) to (e^0.7)^t. Exponents is one of the topics that are covered in pre-calculus, so the ability to convert exponents from one form to another may be what the teacher is looking for here.
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04/08/22

Donald W.

Also, using 2.01, 2.013, 2.0137, or ... doesn't fix the accuracy problem. It just makes the problem smaller, but the problem still exists.
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04/08/22

Donald W.

One could also argue that Q = 0.35 * (2.01....)^t is not a valid answer because it's a different equation that gives different results. You're not converting the equation as the problem is asking, but providing a completely different equation. The only instance when your equation provides the same result is when t = 0. For all other values of t, your equation gives a different result.
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04/08/22

Dayv O.

perhaps you are right about exponent work for student. But it is also eventually important student know Ln2=about .7 (or .69 more accurately). It helps in grasping magnitude of quantities.
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04/08/22

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