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Hi

Can you please help me solve for A in this equation ((1+A)^B)-1xC=D

You have to know logarithm or understand the concept of fractional power to understand the solution. (I will present the solution without logarithm.)

1xC is just C. So your equation is

(1+A)^B - C = 0

add C to both sides of the equation to yield

(1+A)^B = C

Now take fractional power (or Bth root) on both sides (this has some restriction, it is only possible when B is of certain kind and C is not negative when B is even.

(1+A) = C^(1/B)

Now subtract one from both sides to get your A isolated on the left side of the equation. I will leave the last step for you and not complete it.

Calculus

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