Simplifying Logarithm Expressions with Unknown Variables

[teleport]

Homework Statement

Given that

$$\log_{4n} 40\sqrt {3} = \log_{3n} 45$$

find

$$n^3$$

Homework Equations

Logarithm properties

The Attempt at a Solution

I can get an expression for n but looks messy, and suspect there is probably a more compact answer. This is what I did:

$$\dfrac{\log 40\sqrt{3}}{\log 4n} = \dfrac{\log 45}{\log 3n} <br /> <br /> \leftrightarrow \dfrac{\log 40\sqrt{3}}{\log 45} = \dfrac {\log 4n}{\log 3n} <br /> <br /> = \dfrac{\log 4 + \log n}{\log 3 + \log n}$$

and then I solved for log n; but again, my answer is 'messy'.

Thanks for any help.

 

[Dick]

I think the answer is just a little 'messy'. I don't think you can simplify it significantly.

 

[teleport]

I have been told the answer is actually very compact.

 

[Dick]

Now that you mention it, if I numerically evaluate the answer, it is suspiciously close to an integer. Can you guess which one I'm thinking of? I can't figure out how to get there directly though.

 

[Dick]

Ok. Got it. Take the expression you get for log(n) and factor all of the numbers into primes. Now if multiply everything out and simplify, then, lo and behold, the result magically becomes log(N)/3. Where N was that number I was thinking of. Pretty painful - but I still can't think of a more direct way. Plenty of practice with the rules of logarithms there.

 

[moose]

Ok, I got the answer after some very round about math. Right now I'm trying to find a way to do it with less work. I got the answer... n^3 is a two digit integer. If I figure out a way to do this more easily, I will throw you in the right direction.

 

[Dick]

Maybe we are taking this 'don't tell the answer thing' too seriously. I suspect that teleport knows the answer too. But this is getting to be sort of fun. So I'll say that the first letter of the last digit of the two digit number is 'f'. Your turn. Finding this with a lot less work would be a cool thing. But maybe somebody just cooked it to work this way.

 

[HallsofIvy]

If $log_{4n}(40\sqrt{3})= log_{3n}(45)$, then, letting a be that mutual value, $(4n)^a= 40\sqrt{3}$ and $(3n)^a= 45$. Dividing one equation by the other $(4/3)^a= 8\sqrt{3}/9= 2^3/3^{3/2}$. That is, 2^{2a}= 2^3 and 3^a= 3^{3/2}: a= 3/2.

Now we have $(4n)^{3/2}= 40\sqrt{3}$ so [itex](4n)^3= 4^3 n^3= 64 n^3= 1600(3)= 4800$. n^3= 4800/64= 75.$[/itex]

 

[teleport]

Nice! However, I don't think you did well in providing the solution.

Sorry for not replying in a few days. I had a lot of workload in my summer courses. But now its done... for now!