• Jason-Li
In summary, the learning materials states that you need to boil this equation down to: $$\frac {1} {2700} + \frac {1} {3930n^2}$$ however, you are having trouble getting rid of the 10^{-5} and are not sure why.

#### Jason-Li

Homework Statement
Working through some fractions in loop-gain of an oscillator and stuck when comparing my answer to the learning materials...
Relevant Equations
algebra & fractions
So my final equation is:

##\frac {1} {2700} + \frac {1} {3930n^2} + 10^{-5}##

I need to boil this down, the learning materials has the following working, but I can't seem to get it
$$\frac {1} {2700} + \frac {1} {3930n^2} + 10^{-5}$$

$$\frac {3930n^2+2700+2700*3930n^2*10^{-5}} {(2700*3930n^2)}$$

But I have the following:

$$\frac {(3930n^2+2700)*10^{-5}+2700*3930n^2*10^{-5}} {(2700*3930n^2)}$$

Not sure why I have the extra 10^{-5} or how to get rid of it?

Unless the following makes mathematical sense? by making 10^{-5} = 1/ 10^{5}

$$\frac {(3930n^2+2700)*10^{5}+2700*3930n^2} {(2700*3930n^2)*10^5}$$
$$\frac {(3930n^2+2700)+2700*3930n^2*10^{-5}} {(2700*3930n^2)}$$

But the problem is $$\frac {(3930n^2+2700)*10^{5}+2700*3930n^2} {(2700*3930n^2)*10^5} ≠ \frac {(3930n^2+2700)+2700*3930n^2*10^{-5}} {(2700*3930n^2)}$$

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docnet
In general $$\frac 1 a + \frac 1 b + c = \frac{b + a + abc}{ab}$$
So, the book is correct.

I think you are confusing ##10^{-5}## with ##\frac 1 {10^5}## in terms of how you treat it as a fraction.

docnet and Jason-Li
PS you can always set ##n = 1## and put each expression into a calculator or spreadsheet. You'll see that your expression is incorrect.

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docnet and Jason-Li
PeroK said:
In general $$\frac 1 a + \frac 1 b + c = \frac{b + a + abc}{ab}$$
So, the book is correct.

I think you are confusing ##10^{-5}## with ##\frac 1 {10^5}## in terms of how you treat it as a fraction.

PeroK,

I should've simplified it down first like you have e.g. abc...

Thanks again!

docnet
PeroK said:
I think you are confusing ##10^{-5}## with ##\frac 1 {10^5}## in terms of how you treat it as a fraction.
That would be fine, since they are equal, as long as it was done correctly.

docnet
Jason-Li said:
So my final equation is: ##\frac {1} {2700} + \frac {1} {3930n^2} + 10^{-5}##
Nit: That's not an equation. An equation is a statement about the equality of two expressions. An equation will always include at least one = symbol.

WWGD, SammyS and docnet
Mark44 said:
Nit: That's not an equation. An equation is a statement about the equality of two expressions. An equation will always include at least one = symbol.
And a good way of remembering is looking at the first 4 letters of the word : Edit :equa(l)(ity)

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WWGD said:
And a good way of remembering is lo9king at the first 4 letters of the word : aqua(l)(ity)
aqua?

Mark44 said:
aqua?
Auto (in)correct strikes again. Let me edit.