# Rearranging of an equation

• iScience
In summary: W = -A_1 B_1^c \frac{1}{-c+1}\left(B_2^{-c+1} - B_1^{-c + 1}\right)= -\frac{A_1 B_1^c}{c-1}\left(B_2^{-1-c} - B_1^{-1-c}\right)

#### iScience

i have a given equation of the following.

W= -A1B1c[$\frac{1}{-c+1}$B-c+1] where the part in brackets is the integrated function of B going from B1 to B2.

then this equation somehow ends up like so

W=$\frac{A_1B_1}{c-1}${((B1)/(B2))^(c-1)-1}

(sorry for the messy format on the right hand side i couldn't get latex to work..)

i don't know how they got from the first equation to the second equation. i don't know where to start because i don't know how certain terms were combined.

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iScience said:
i have a given equation of the following.

W= -A1B1c[$\frac{1}{-c+1}$B-c+1] where the part in brackets is the integrated function of B going from B1 to B2.
Is this what you mean?

$$W = -A_1B_1^c \int_{B_1}^{B_2} \frac{B^{-c + 1} dB}{-c + 1}$$
iScience said:
then this equation somehow ends up like so

W=$\frac{A_1B_1}{c-1}${((B1)/(B2))^(c-1)-1}

(sorry for the messy format on the right hand side i couldn't get latex to work..)

i don't know how they got from the first equation to the second equation. i don't know where to start because i don't know how certain terms were combined.

If you right-click on the integral I wrote, you can see the LaTeX that creates it.

thanks, however that wasn't what i meant. that function inside the bracket is 'already' integrated; what i was trying to say was that the limits just weren't taken.

W = $-A_1B_1^c \frac{B^{-c + 1}}{-c + 1}$

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Are you sure that your antiderivative is correct? I'm thinking you might have made a mistake. What was the problem you started with?

iScience said:
i have a given equation of the following.

W= -A1B1c[$\frac{1}{-c+1}$B-c+1] where the part in brackets is the integrated function of B going from B1 to B2.

So that's
$$W = -A_1 B_1^c \frac{1}{-c+1}\left(B_2^{-c+1} - B_1^{-c + 1}\right) = \frac{A_1 B_1^c }{c-1}\left(B_2^{1-c} - B_1^{1-c}\right)$$
after tidying up some signs.

Now we pull a common factor of $B_1^{1-c}$ out of the bracket:
$$W = \frac{A_1 B_1^c B_1^{1-c}}{c - 1} \left( \frac{B_2^{1-c}}{B_1^{1-c}} - 1\right) = \frac{A_1 B_1}{c - 1}\left( \left(\frac{B_2}{B_1}\right)^{1-c} - 1\right)$$

Finally we flip the fraction in the bracket, remembering to multiply the exponent by -1 as we do so.

## What is the purpose of rearranging an equation?

Rearranging an equation allows us to solve for a different variable or to simplify the equation in order to make it easier to work with.

## How do you determine which variable to solve for when rearranging an equation?

This depends on the specific problem you are trying to solve. Look at the given information and what you are trying to find, and choose the variable that will help you get to your desired solution.

## What are the steps for rearranging an equation?

1. Identify the variable you want to solve for.2. Isolate that variable on one side of the equation by moving all other terms to the other side.3. Simplify the equation, if necessary.4. Check your solution by plugging it back into the original equation.

## What are some common mistakes to avoid when rearranging an equation?

Some common mistakes include:- Not correctly distributing a negative sign when moving terms to the other side of the equation- Forgetting to apply the same operation to both sides of the equation- Making a calculation error during the simplification step- Not checking the solution by plugging it back into the original equation

## How can rearranging an equation be useful in real-world applications?

Rearranging an equation can help us solve practical problems, such as finding the missing variable in a scientific or mathematical formula. It can also be useful in understanding relationships between different quantities and predicting how they will change in different scenarios.