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.

The growth rate of the function $$f(x) = b a ^ x$$ is $17\%$, and $f(0) = 24$

What I am trying to figure out is how to find out what $a$ and $b$ in this equation are?

share|improve this question
    
What happens when you substitute $x=0$ into the given expression for $f(x)$? –  Brad Nov 21 '12 at 18:52
    
Do you know some calculus? –  André Nicolas Nov 21 '12 at 18:52
    
As it says, f(0) = 24 - all that i know is above –  Frederik Witte Nov 21 '12 at 18:55
    
Hint 1: f(0) = b a^x = b a^0 = b = 24. Is 17 the value at some given x or for some value for 'a'? –  Amzoti Nov 21 '12 at 18:55
    
17% is the growth of the function, it's expotential, it's raising with 17% for each time it moves along the x axis –  Frederik Witte Nov 21 '12 at 18:56

2 Answers 2

up vote 5 down vote accepted

In your case you are given $f(0) = 24$. This gives us that $$b a^0 = 24 \implies b = 24$$

You are also given that the growth rate is $0.17$.

Growth rate is typically defined as $$\dfrac1f\dfrac{df}{dx}$$ Since $f(x) = 24 a^x$, we have that $\dfrac{df}{dx} = 24 a^x \log(a)$. Hence, growth rate is $$\dfrac1f\dfrac{df}{dx} = \dfrac{24 a^x \log(a)}{24 a^x} = 0.17$$ This gives us that $a = e^{0.17}$. Hence, $$f(x) = 24e^{0.17x}$$

share|improve this answer
3  
genius, master of math - thanks alot! –  Frederik Witte Nov 21 '12 at 18:58

First use the fact that $f(0)=24$: since $f(x)=ba^x$, $24=f(0)=ba^0=b\cdot1=b$, and we now know that $f=24a^x$. Now we use the $17$% growth rate to determine $a$: to get an increase of $17$% with each unit increase in $x$, you need to multiply by $1.17$ ($117$%) every time $x$ increases by $1$, so $a=1.17$, and $f(x)=24(1.17^x)$.

share|improve this answer

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.