Issues with an exponential problem

[Cmunro]

Homework Statement

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.

Homework 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)

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

 

[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.

 

[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?

 

[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.

 

[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?

 

[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

 

[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.

 

[HallsofIvy]

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

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.

 

[theperthvan]

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.

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.

 

[Cmunro]

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.

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!