Melody

We need someone better than me at geometry proofs here   

Any takers  ??

Thios is the diagram. 

The aim is to show that the 3 red lines are equal in length.

OR

\(log_{(4/9)}(\frac{27}{8})^{3/4}\\ =log_{(4/9)}(\frac{3}{2})^{9/4}\\ =log_{(4/9)}(\frac{2}{3})^{-9/4}\\ =log_{(4/9)}(\frac{4}{9})^{-9/8}\\ =\frac{-9}{8}\)

Let

\(y=log_{(4/9)}(27/8)^{(3/4)}\\ y= \frac{log(27/8)^{(3/4)}}{log(4/9)}\\ y= \frac{(3/4)log(27/8)}{log(2/3)^2}\\ y= \frac{(3/4)log(3/2)^3}{log(2/3)^2}\\ y= \frac{(9/4)log(3/2)}{2log(2/3)}\\ y= \frac{9log(3/2)}{8log(2/3)}\\ y= \frac{9log(3/2)}{8log(3/2)^{-1}}\\ y= \frac{9log(3/2)}{-8log(3/2)}\\ y=\frac{-9}{8} \)

Invisable :))

inverse tan is the same as arc tan so type in    atan(number)

The most important thing is that you know what it looks like.

Here are some pictures. You certainly will not be able to see it if you do not know what you are supposed to be lookng for. :)

yea, thanks Ghosty :)

There are no questions here!

If a and b must be positive integers, what is the largest integer n such that 13a + 18b = n has no solutions?

I I don't know, I am getting really confused here . ://

This is what I am considering.

I used the euclidean algorithm, followed by the extended euclidean algorithm to come up with the generic solutions of

a= 7n-18k

b= -5n+13k

Where k is any integer.  So they are the answers to the diaphantine equation.   13a+18b=n

since a and be must both be positive

\(7n-18k>0\\ n>\frac{18k}{7}\qquad \qquad n>\frac{90k}{35} \\ -5n+13k>0\\ n<\frac{13k}{5}\qquad \qquad n<\frac{91k}{35} \\~\\ \frac{90k}{35} < n<\frac{91k}{35} \)

Mmm...   I have no idea where this is going. 

Yes, I know, Tuesday morning I think. :)