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label Calculus
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schedule 1 Day
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Jul 2nd, 2015

Thank you for the opportunity to help you with your question! the solution is as such:

y= ab^x

for 1st point P  12= ab^ -2

for 2nd point Q 3/8 = ab ^3

on dividing the equations we get b= 1/2 and a = 3

now the tangent is parallel to PQ. so slope of tangent on the curve at the  required point is = slope of PQ. /

slope of PQ= (y2- y1) / (x2- x1)

                   = - 2.325

now slope of tangent on the curve at the  required point is -2.325

but we know slope of tangent is =dy/dx

so dy/dx =   d(ab^x)/dx 

              =  y lnb

now at (x,y)

the slope= -2.325

so y= 3.354

putting in the equation when y= 3.354 then x= 0.07

Please let me know if you need any clarification. I'm always happy to answer your questions.awaiting a positive feedback. It was lovely helping you. Looking forward to help you again. A positive feedback awaited thank you.
Jul 2nd, 2015

sorry x= -0.16

Jul 2nd, 2015

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