Math in the thin lens equation

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Main Question or Discussion Point

Why isn't 1/f=1/do+1/di just the mathematical equivalent to f=do+di? Can't you raise all terms in the equation to the power of negative 1 to get the latter equation? The maths of reciprocals is confusing me lol.

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That is not the lens eqn.
The lens equation is :- 1/f = 1/v - 1/u
Where v:- image distance with sign convention
u:- object distance with sign convention
just take reciprocal of whole eqn

$\frac {1}{\frac{1}{f}} = \frac {1}{\frac{1}{v} - \frac{1}{u}}$

$f = \frac{1}{\frac{(u-v)}{uv}}$

$f= \frac{uv}{u-v}$

Clearly it is self explanatory.

That is not the lens eqn.
The lens equation is :- 1/f = 1/v - 1/u
Kuzon has written a perfectly acceptable version of the lens equation, one that is commonly used.

Kuzon: raising to a power does not follow a distributive law as does multiplication. So we have a(b + c) = ab + ac but (b + c)2 ≠ b2 + c2.

Kuzon has written a perfectly acceptable version of the lens equation, one that is commonly used.

Kuzon: raising to a power does not follow a distributive law as does multiplication. So we have a(b + c) = ab + ac but (b + c)2 ≠ b2 + c2.
Sorry, I did not know that version of lens equation; as in our country the lens equation version used is different.

I have acquired the same as @Sahil Kukreja . The relation is about their reciprocals, and cannot directly be changed to what @Kuzon wrote.