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} $

1 Answer

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.