Rearranging formula using logarithms

In summary, the conversation involved the exchange of ideas to help the speaker understand how to use logarithms to solve a given formula. The speaker had a homework problem involving finding the current and time in a circuit, and had attempted to solve it using a rearranged equation but was getting incorrect answers. The expert summary provided guidance on using the ln button instead of the log button on a calculator, which led to the correct solution. The speaker expressed gratitude for the help and blamed their tutor for not explaining the use of the ln button.
  • #1
Mr_Mole
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3
Homework Statement
I have a homework where I need to use logarithms to make an unknown the subject of a given formula. I know the answer I need (different values in the formula till one worked) but I can’t show how it’s done using logs.

If the flow of current (i), Amperes, in a capacitor at time (t), seconds, increases at an exponential according to the equation below:
i=6.7(1-e^(-t⁄RC) )
The resistance of the circuit is 28kΩ and the capacitance is 15µF.
Calculate the current after 750ms
Calculate the time for the current to reach 5A.
Relevant Equations
i=6.7(1-e^(-t⁄RC) )
My attempt that doesn’t work. t=-RC ln(1-I/6.7)
 
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  • #2
Your "attempt that doesn't work" looks correct to me. To show why this is true via logarithms you'll need to use the idea that ln(e^x)=x for x>0. In other words, make e^x the subject of the equation and then take the natural log of both sides.
 
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  • #3
It looks good to me also. Why do you think it's not correct?
 
  • #4
What is the answer supposed to be?
 
  • #5
DaveE said:
It looks good to me also. Why do you think it's not correct?
Because when I put values into it then I get the wrong answer. Using the information given in the question I get an answer of 5.576A for the first part. Putting that into the rearranged formula gives me a t value of 326milliseconds instead of 750milliseconds.
 
  • #6
Mr_Mole said:
Because when I put values into it then I get the wrong answer. Using the information given in the question I get an answer of 5.576A for the first part. Putting that into the rearranged formula gives me a t value of 326milliseconds instead of 750milliseconds.
How are you getting 326ms?
 
  • #7
DaveE said:
It looks good to me also. Why do you think it's not correct?
Mr_Mole said:
Homework Statement: I have a homework where I need to use logarithms to make an unknown the subject of a given formula. I know the answer I need (different values in the formula till one worked) but I can’t show how it’s done using logs.

If the flow of current (i), Amperes, in a capacitor at time (t), seconds, increases at an exponential according to the equation below:
i=6.7(1-e^(-t⁄RC) )
The resistance of the circuit is 28kΩ and the capacitance is 15µF.
Calculate the current after 750ms
Calculate the time for the current to reach 5A.
Relevant Equations: i=6.7(1-e^(-t⁄RC) )

My attempt that doesn’t work. t=-RC ln(1-I/6.7)
I’m not expecting the current to be 5A at 750ms. When rearranged to make t the subject, I would expect t to be 750ms when I input the current obtained when i is the subject. That was around 5.6A. I’m getting 325ms out which led me to believe it was incorrect. This aspect of maths baffles me hence how I ended up here relying on the kindness of helpful strangers with bigger brains than I.
 
  • #8
Mr_Mole said:
I’m not expecting the current to be 5A at 750ms. When rearranged to make t the subject, I would expect t to be 750ms when I input the current obtained when i is the subject. That was around 5.6A. I’m getting 325ms out which led me to believe it was incorrect. This aspect of maths baffles me hence how I ended up here relying on the kindness of helpful strangers with bigger brains than I.
Can you explain how you are getting times of 325ms or 326ms?
 
  • #9
Mr_Mole said:
I’m getting 325ms out
One of the more frequent mishaps: you used LOG on your calculator instead of LN

:smile:

##\ ##
 
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  • #10
YouAreAwesome said:
Can you explain how you are getting times of 325ms or 326ms?
I input the values into the rearranged equation. Using the same values given and the original answer I would expect to get the original i value returned to prove I had rearranged correctly. Unless I’m missing a step somewhere in the process and doing the rearranged equation incorrectly.
 
  • #11
BvU said:
One of the more frequent mishaps: you used LOG on your calculator instead of LN

:smile:

##\ ##
We have a winner! If only the maths tutor had explained that. No mention of an ln button. It comes out correct now. You are my hero! It was driving me mad last night as I couldn’t see the error. I will sleep well tonight!
 
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  • #12
Mr_Mole said:
I input the values into the rearranged equation. Using the same values given and the original answer I would expect to get the original i value returned to prove I had rearranged correctly. Unless I’m missing a step somewhere in the process and doing the rearranged equation incorrectly.
Thank you for helping. It’s been pointed out I didn’t know the existence of the ln button. I blame the tutor for not explain that one.
 

FAQ: Rearranging formula using logarithms

What is the purpose of using logarithms to rearrange formulas?

Logarithms are used to simplify equations, particularly those involving exponential functions. They help transform multiplicative relationships into additive ones, making it easier to solve for unknown variables.

How do you apply logarithms to both sides of an equation?

To apply logarithms to both sides of an equation, you take the logarithm of each side. For example, if you have the equation \(a = b^c\), you can take the natural logarithm (ln) of both sides to get \(\ln(a) = \ln(b^c)\). Using the properties of logarithms, this can be further simplified to \(\ln(a) = c \cdot \ln(b)\).

What are the common logarithmic properties used in rearranging formulas?

The common logarithmic properties include:1. \(\log(ab) = \log(a) + \log(b)\) (Product Rule)2. \(\log\left(\frac{a}{b}\right) = \log(a) - \log(b)\) (Quotient Rule)3. \(\log(a^b) = b \cdot \log(a)\) (Power Rule)

Can logarithms be used with any base, and how does the base affect the process?

Yes, logarithms can be used with any base, such as base 10 (common logarithm) or base \(e\) (natural logarithm). The base affects the process in that the choice of base should be consistent throughout the equation. The natural logarithm (ln) is often preferred in scientific contexts due to its relationship with the exponential function \(e\).

How do you solve for an exponent using logarithms?

To solve for an exponent using logarithms, you first isolate the exponential expression. For example, if you have \(y = a^x\), take the logarithm of both sides: \(\log(y) = \log(a^x)\). Using the power rule, this becomes \(\log(y) = x \cdot \log(a)\). Finally, solve for \(x\) by dividing both sides by \(\log(a)\): \(x = \frac{\log(y)}{\log(a)}\).

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