# 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. $f(x) = c \cdot x^r$

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

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

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

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

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

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

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

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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