point of intersection

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June 18th 2010, 03:44 PM
Hellooo

point of intersection

How the find the point of intersection of the two functions: y=300(1.05)^x and y=1000(0.92)^x? Help, Thanks(Worried)
June 18th 2010, 03:58 PM
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Quote:

Originally Posted by Hellooo

How the find the point of intersection of the two functions: y=300(1.05)^x and y=1000(0.92)^x? Help, Thanks(Worried)

I haven't looked to ensure that there exists a unique intersection point, but you can find any intersection points by solving

300(1.05)^x = 1000(0.92)^x
June 18th 2010, 07:43 PM
Bacterius

Hello,
you can let a = 1.05 and b = 0.92 to make it simpler, then you have to solve :

300 a^x = 1000 b^x

Taking the logarithms of base "a" yields :

log_a(300 a^x) = log_a(1000 b^x)

Simplifying using the log laws :

log_a(300) + log_a(a^x) = log_a(1000) + log_a(b^x)

Keep simplifying :

log_a(300) + x = log_a(1000) + x log_a(b)

Isolate the x's on one side :

log_a(300) - log_a(1000) = x log_a(b) - x

Simplify :

log_a(300 / 1000) = x(log_a(b) - 1)

Alright :

(log_a(0.3)) / (log_a(b) - 1) = x

Now replace a and b with the values defined at the start, do the calculation and you got x (provided it exists, you'll know it doesn't if you get a math error :) )

EDIT : sorry, used 100 instead of 1000, corrected and checked against Wolfram
June 18th 2010, 08:28 PM
simplependulum

$300(1.05)^x = 1000(0.92)^x$

$(1.05)^x / (0.92)^x = 10/3 <br />$
$(1.05/0.92) ^x = 10/3$

$x log(1.05/0.92) = log(10/3)$

$x = log(10/3) / log(1.05/0.92)$

$= 9.109152249....$
June 19th 2010, 07:31 AM
Hellooo

Thank you very much!