# Issues with an exponential problem

1. Jan 5, 2007

### Cmunro

1. The problem statement, all variables and given/known data
I have been told: Two variables x and n are connected by the formula
x=a( n^b). When n = 2, x = 37; when n = 3, x = 66. I have been told to find a and b.

2. Relevant equations

x=a(n^b) (I've put the brackets there because I don't know how to differenciate between times and "x" --below the x inbetween is a times not an x)

3. The attempt at a solution

37=a x 2^b 66= a x 3^b or 66/3^b =a

37 = (66/3^b) x 2^b
37= (66 x 2^b)/3^b
log37=(log66 x blog2)/blog3

Ok so I've gotten to this point - but I don't know how to get b on its own! Can anyone give me a hand here?

Thanks,
Cat

2. Jan 6, 2007

### gabee

Here's a hint on how I solved it:

Since $$x = an^b$$, rearrange the equation to read $$a = \frac{x}{n^b}$$. Find a system of equations by plugging in the numbers for x and n given to you in the problem. You can then equate these two by a = a. See if you can rearrange the equation into a format that will allow you to then put it into log form and solve for the exponent b.

3. Jan 6, 2007

### Cmunro

ok, so I've tried it but I've ended up with the same problem. I don't know how to move b over - without cancelling out b.

as a=37/2^b and a=66/3^b
if a=a then: log37/blog2 =log66/blog3

can I flip the base or something? I'm not a naturally good mathematician, so sometimes I can miss the obvious that someone who is can see. What is there that I'm missing?

4. Jan 6, 2007

### theperthvan

Divide one equation by the other, getting rid of a.

Then you can find b easily.

Sub b back into an equation to find a.

5. Jan 6, 2007

### Cmunro

Ok I must be really thick, but I really don't see this at all.

(66=a x 3^b) / (37 = a x 2^b)

66/37 =(3^b)/(2^b)
log 66/37 = b log3/ b log 2

essentually this cancels b out completely - right?

6. Jan 6, 2007

### theperthvan

up to here is good

66/37 =(3^b)/(2^b)

and then what does (3^b)/(2^b) become

HINT: (a^m)/(b^m) = (a/b)^m

7. Jan 6, 2007

### Cmunro

ok!!! I get this now! so then

66/37 = (3/2)^b then I log both sides and voila!

Thanks I really appreciate it! I haven't actually seen that rule before, but this is useful to know.

8. Jan 6, 2007

### HallsofIvy

Staff Emeritus
Your error is that log(x/y) is NOT log(x)/log(y), it is log(x)- log(y).
You have log(66/37)= b log(3)- b log(2)= b(log(3)- log(2))

b= (log(66)- log(37))/(log(3)- log(2))

Of course, that's exactly what you get solving (3/2)b= 66/37.

9. Jan 6, 2007

### theperthvan

No worries. That's quite a fundamental rule, so maybe you should review the Indice Laws. They're pretty easy and become second nature.
Also, from what HallsofIvy pointed out, maybe review your log laws too.

10. Jan 6, 2007

### Cmunro

Ahh I see exactly what you mean. Back to reviewing the log rules! I'm trying to revise all these things at the moment you see, but a lot of it has flown out of my head. Anyway, important errors to learn from - so thank you for pointing this out!