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February 1, 2015

February 1, 2015

Posted by **Marlanea** on Monday, October 1, 2012 at 9:45pm.

- Calculus -
**Reiny**, Monday, October 1, 2012 at 9:56pmdy/dx = 2.5(5^x)ln5

when x=0 from (0, 2.5)

dy/dx = 2.5ln5

so tangent is y= 2.5ln5 x + c

at (0 , 2.5)

2.5 = 0 + c

so the equation is y = 2.5ln5 x + 2.5

at the x-axis, y =0

0 = 2.5ln5 x + 2.5

2.5ln5 x = -2.5

ln5 x = -1

x = -1/ln5 or appr -.62

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