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April 16th, 2018, 09:49 AM   #1
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Solve for x

y=a*(1+(b*m*x))^(-1/b)

How to solve for x?
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April 16th, 2018, 09:57 AM   #2
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Why is this in calculus?

Here's a start (making certain assumptions on the values of a, b, and m):

$y = a(1 + bmx)^{(-1/b)} \implies y = \dfrac{a}{\sqrt[b]{1 + bmx}} \implies$

$\sqrt[b]{1 + bmx} = \dfrac{a}{y} \implies WHAT?$
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April 16th, 2018, 10:13 AM   #3
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y / a = 1 / (1 + bmx)^(1 / b)

(1 + bmx)^(1 / b) = a / y

Can you finish it?
We do not give out full solutions.
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April 16th, 2018, 10:26 AM   #4
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Moved from Calculus to Algebra.
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April 16th, 2018, 10:35 AM   #5
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ok. thanks
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April 16th, 2018, 11:17 AM   #6
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Hint: if x^a = y, then x = y^(1 / a)
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April 16th, 2018, 02:19 PM   #7
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x=((((y/a)^(-b))-1)/bm)

Did I get it right?
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April 16th, 2018, 03:24 PM   #8
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Quote:
Originally Posted by amirriazahmed View Post
x=((((y/a)^(-b))-1)/bm)
Did I get it right?
YES...but you need to bracket the denominator bm:
x=((((y/a)^(-b))-1)/(bm))

You have extra brackets (ok to leave as they are but not required);
this is sufficient: x=((y / a)^(-b) - 1) / (bm)

Can be rearranged this way:
x = ((a / y)^b - 1) / (bm)

You can check if correct by using original equation and assigning
values to a,b,m,x then calculating y from these values,
then seeing if x comes out ok using your "solution equation".
Try it; say a=2, b=3, c=4, x=5.
See what I mean?
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April 16th, 2018, 03:27 PM   #9
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Nearly... x = ((y/a)^(-b) - 1)/(bm) has correct use of parentheses,
or you could give x = ((a/y)^b - 1)/(bm).
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