So with the equation MU=ln(C1)+ln(160-1.1C1) the derivative is

[surferbarney0729]

So with the equation MU=ln(C1)+ln(160-1.1C1)

the derivative is 1/C1 - 1.1/(160-1.1C1)

How does one solve and identify the derivative instead being seen and written as

2.2C1 = 160

What are the steps taken to get to the latter derivative. the second format was much easier to solve the final equation, but I am not sure how to 2.2C1 = 160 was found

Any help?

 

[mathman]

The equation you want is from 1/C1 - 1.1/(160-1.1C1) = 0

 

[HallsofIvy]

So with the equation MU=ln(C1)+ln(160-1.1C1)

the derivative is 1/C1 - 1.1/(160-1.1C1)

How does one solve and identify the derivative instead being seen and written as

2.2C1 = 160

What are the steps taken to get to the latter derivative. the second format was much easier to solve the final equation, but I am not sure how to 2.2C1 = 160 was found

Any help?

The "second format" cannot be derived from the first without additional information. I presume that the actual problem was to find where the derivative is equal to 0: 1/C1- 1.1/(160- 1.1C1)= 0. Add 1.1/(160- 1.1C1) to both sides to get
1/C1= 1.1/(160- 1.1C1). Now multiply both sides by C1 and 160- 1.1C1 to get rid of the fraction: 160- 1.1C1= 1.1C1. Adding 1.1C1 to both sides gives 160= 2.2C1.