Minimum value of logarithmic equation

1. Jun 29, 2008

ritwik06

1. The problem statement, all variables and given/known data

For x>1, find the minimum possible value of $$2 log_{2}x-log_{x}(0.01)$$

3. The attempt at a solution
Greater the value of x, greater is the value of expression. Right?
I tried to differentiate it, but it was no help. The derivative becomes zero when |log x|=$$\sqrt{log 2}$$

Help me further.

2. Jun 29, 2008

Defennder

Re: Log

What did you get when you differentiate it?

3. Jun 29, 2008

ritwik06

Re: Log

Why do ya ask that? Is it wrong?
I got:
2*(1/x ln 2)-2*(1/log^{2} x)*(1/(x ln 10))

4. Jun 29, 2008

Defennder

Re: Log

Well, apparently I can't tell if it's correct unless I know your working. I can't read what you wrote that. Is it $$\frac{2 \ln x}{\ln 2} - \frac{2}{\log_2 x} \left( \frac{1}{x \ln 10} \right)$$.

If so then I don't think it's correct.

5. Jun 30, 2008

ritwik06

Re: Log

$$\frac{2}{x ln 2}-\frac{2}{(log^{2} x)*(x ln 10)}$$

6. Jun 30, 2008

dynamicsolo

Re: Log

This problem sets a similar trap to that of another problem you asked about. It would make life easier if you rewrote the $$log_{x}(0.01)$$ term as a log-base-2 term first, so you could combine it with the first term...

(You could grind through the differentiation you have, but it is a rather cumbersome "hammer" to use on the problem.)

7. Jun 30, 2008

happyg1

Re: Log

I got:
$$\frac{2}{x ln 2}-\frac{2 ln 10}{x ln^2 x}$$
for the derivative.
I set it equal to zero, cross multiplied and came up with:

$$ln^2x=ln 10 ln 2$$

Dunno if that helps...your derivative was different than mine.
CC

Last edited: Jun 30, 2008
8. Jun 30, 2008

ritwik06

Re: Log

rewriting:
$$2log_{2}x+\frac{2(1+log_{2}5)}{log_{2}x}$$

9. Jun 30, 2008

happyg1

Re: Log

Take the square root of both sides, ignore the absolute value bars, because for x>1 the thing is positive, then take the exponential of both sides. I got x=3.53722.......
plug that back in to get the y value.
CC

Last edited: Jun 30, 2008