# Exponential and power functions through two points

I have a problem where I'm asked to determine the constants of exponential and power functions that go through both points (5, 50) and (10, 1600). I have tried to solve them below, but would appreciate it if someone could check. Would also appreciate feedback on how I could optimize my notation, if anyone has any thoughts on that.

Exponential, i.e. $f(x) = c \cdot a^x$

$50 = c \cdot a^5 \\ 1600 = c \cdot a^{10}$

$ln(50) = ln( c ) + 5ln(a) \\ ln(1600) = ln( c ) + 10ln(a)$

$ln(50)-ln(1600) = 5ln(a) - 10ln(a) \Rightarrow ln(\frac{50}{1600}) = -5ln(a) \Rightarrow ln(\frac{1}{32}) = -5ln(a) \Rightarrow -ln(32) = -5ln(a) \Rightarrow \frac{-ln(32)}{-5} = ln(a) \Rightarrow e^{ln(a)} = e^{\frac{ln(32)}{5}} = 2$

$f(x) = c \cdot 2^{x} \Rightarrow 50 = c \cdot 2^{5}, 1600 = c \cdot 2^{10}$

$50 = 32c \Rightarrow c = \frac{50}{32} = \frac{25}{16}$

$1600 = 1024c \Rightarrow c = \frac{1600}{1024} = \frac{25}{16}$

$f(x) = \frac{25}{16}2^{x}$

Power, i.e. $f(x) = c \cdot x^r$

$50 = c \cdot 5^r \\ 1600 = c \cdot 10^r$

$ln(50) = ln( c ) + rln(5) \\ ln(1600) = ln( c ) + rln(10)$

$ln(50) - ln(1600) = r(ln(5) - ln(10))$

$\frac{ln(50) - ln(1600)}{ln(5) - ln(10)} = r \Rightarrow r = 5$

$50 = c \cdot 5^5 \Rightarrow c = \frac{2}{125}$

$1600 = c \cdot 10^5 \Rightarrow c = \frac{2}{125}$

$f(x) = \frac{2}{125}x^{5}$

-
these look correct. what do you mean by optimize your notation? you can remove the extra step when evaluating c, you only need to do it once. – mythealias Nov 11 '12 at 14:08
Thank you for checking. Sorry I didn't mean the notation but if there were any actual steps that I could have skipped, which you both helped with so thanks. :) – jiku Nov 11 '12 at 22:10

The steps seem to be good. In both cases, you could divide your first equation by the second one (or vice versa) and then take ln on both sides. It would save you some time.

-
Thanks for checking and the tip. :) – jiku Nov 11 '12 at 22:10