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)

Printable View

- Jun 18th 2010, 03:44 PMHellooopoint 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)

- Jun 18th 2010, 03:58 PMundefined
- Jun 18th 2010, 07:43 PMBacterius
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 - Jun 18th 2010, 08:28 PMsimplependulum
$\displaystyle 300(1.05)^x = 1000(0.92)^x $

$\displaystyle (1.05)^x / (0.92)^x = 10/3

$

$\displaystyle (1.05/0.92) ^x = 10/3 $

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

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

$\displaystyle = 9.109152249....$ - Jun 19th 2010, 07:31 AMHellooo
Thank you very much!