Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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} $

share|improve this question
    
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. :) –  eee3 Nov 11 '12 at 22:10
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
Thanks for checking and the tip. :) –  eee3 Nov 11 '12 at 22:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.