# Logarithm Problem I can't figure out

• Nov 3rd 2009, 05:08 PM
ConMan
Logarithm Problem I can't figure out
These questions are from several math contests, and I'm trying to do these but I can't figure it out.

1. If x and y > 0 , log(base y)x + log(base x)y = 10/3 ,and (x)(y)=144
,find x + y /2

2. How many real numbers x satisfy the equation (1/5) log(base 2)x = sin(5pi x) ?

I tried making y=144/x and plugging that into the equation, but I have no idea what to do for the second question

Any help from you guys is greatly appreciated :)
• Nov 4th 2009, 12:26 AM
earboth
Quote:

Originally Posted by ConMan
These questions are from several math contests, and I'm trying to do these but I can't figure it out.

1. If x and y > 0 , log(base y)x + log(base x)y = 10/3 ,and (x)(y)=144
,find x + y /2

...

1. You are supposed to know that $\displaystyle \log_b(a)=\dfrac1{\log_a(b)}$

2. Using this rule your equation becomes:

$\displaystyle \log_y(x)+\dfrac1{\log_y(x)} = \dfrac{10}3$

Use the substitution $\displaystyle z = \log_y(x)$ . Then you have:

$\displaystyle z-\dfrac1z = \dfrac{10}3~\implies~z^2-\dfrac{10}3 z + 1 = 0$

which yields $\displaystyle z = 3~\vee~z = \dfrac13$

3. Re-substituting yields:

$\displaystyle x = y^3~\vee~x = \sqrt[3]{y}$

4. Now use the second equation $\displaystyle x\cdot y = 144$ to calculate the values of x and y.
• Nov 4th 2009, 02:08 PM
ConMan