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 Algebra Pre-Algebra and Basic Algebra Math Forum

 April 16th, 2018, 09:49 AM #1 Newbie   Joined: Apr 2018 From: Calgary Posts: 3 Thanks: 0 Solve for x y=a*(1+(b*m*x))^(-1/b) How to solve for x? April 16th, 2018, 09:57 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 550 Why is this in calculus? Here's a start (making certain assumptions on the values of a, b, and m): $y = a(1 + bmx)^{(-1/b)} \implies y = \dfrac{a}{\sqrt[b]{1 + bmx}} \implies$ $\sqrt[b]{1 + bmx} = \dfrac{a}{y} \implies WHAT?$ April 16th, 2018, 10:13 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,145 Thanks: 1003 y / a = 1 / (1 + bmx)^(1 / b) (1 + bmx)^(1 / b) = a / y Can you finish it? We do not give out full solutions. April 16th, 2018, 10:26 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,383 Thanks: 2011 Moved from Calculus to Algebra. April 16th, 2018, 10:35 AM #5 Newbie   Joined: Apr 2018 From: Calgary Posts: 3 Thanks: 0 ok. thanks April 16th, 2018, 11:17 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,145 Thanks: 1003 Hint: if x^a = y, then x = y^(1 / a) April 16th, 2018, 02:19 PM #7 Newbie   Joined: Apr 2018 From: Calgary Posts: 3 Thanks: 0 x=((((y/a)^(-b))-1)/bm) Did I get it right? April 16th, 2018, 03:24 PM   #8
Math Team

Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,145
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Quote:
 Originally Posted by amirriazahmed 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? April 16th, 2018, 03:27 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,383 Thanks: 2011 Nearly... x = ((y/a)^(-b) - 1)/(bm) has correct use of parentheses, or you could give x = ((a/y)^b - 1)/(bm). Tags solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gen_shao Algebra 12 November 2nd, 2014 06:11 AM

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