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.
 

What is the purpose of rearranging formulas using logarithms?

Rearranging formulas using logarithms is a useful technique in mathematics and science for solving equations that involve exponential or logarithmic functions. It allows us to isolate a specific variable and solve for its value, even when it is part of a complex equation.

How do logarithms help in rearranging formulas?

Logarithms are useful because they allow us to convert exponential equations into linear equations, which are easier to manipulate. By taking the logarithm of both sides of an equation, we can bring the variable out of the exponent and solve for its value.

What are the steps for rearranging a formula using logarithms?

The steps for rearranging a formula using logarithms are as follows:1. Identify the variable that you want to isolate.2. Take the logarithm of both sides of the equation.3. Use the properties of logarithms to simplify the equation.4. Solve for the isolated variable.5. Check your solution by plugging it back into the original equation.

Can logarithms be used to solve any type of equation?

No, logarithms are most useful for solving equations that involve exponential or logarithmic functions. They cannot be used to solve equations that do not contain these types of functions.

Are there any limitations to using logarithms for rearranging formulas?

One limitation of using logarithms is that they can only be used for positive values. This means that any solutions obtained using logarithms must be verified to ensure they are valid for the given equation. Additionally, logarithms may not always be the most efficient method for solving a particular equation, so it is important to consider other techniques as well.

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