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- Dec 22nd 2007, 09:39 AMperashsolve inequality
- Dec 22nd 2007, 11:00 AMticbol
Log(base x)[5/2 -1/x] > 1

[5/2 -1/x] > x^1

2.5 -1/x > x

Multiply both sides by x,

2.5x -1 > x^2 --------no change in sense of inequality since x is positive

x^2 -2.5x +1 < 0

2x^2 -5x +2 < 0

(2x -1)(x -2) < 0

x < 1/2 -----------(1)

or x < 2 -----------(2)

And, since there are no real logs of negative numbers, then,

5/2 -1/x > 0

5x -2 > 0

x > 2/5

x > 0.4 ------------(3)

Investigate those three critical points.

i) In the interval (0.4, 1/2)

say x = 0.45,

Log(base 0.45)[5/2 -1/(0.45)] >? 1

0.2778 >? (0.45)^1

No, so this interval is not a solution.

ii) In the interval (1/2, 2)

say x = 1.5,

Log(base 1.5)[5/2 -1/(1.5)] >? 1

1.8333 >? (1.5)^1

Yes, so this interval is a solution.

iii) In the interval (2, infinity)

say x = 5,

Log(base 5)[5/2 -1/5)] >? 1

2.3 >? (5)^1

No, so this interval is not a solution.

Therefore, x = (1/2, 2), or 1/2 < x < 2 ------------answer. - Dec 22nd 2007, 11:01 AMtopsquark
Well, for starters to have valid argument for the logarithm we require

so we must ensure that this condition holds.

Now,

by the change of base formula. (I'm using base 10 here, but feel free to use the base of your choice. It doesn't affect the answer.)

Since log(x) is never negative:

Again, since log( ) is never negative, and the function is one to one:

Can you take this from here?

-Dan - Dec 22nd 2007, 11:12 AMred_dog
First of all we have to put the conditions for the existance of logarithm and for the base:

and

If then

.

If .

Then - Dec 22nd 2007, 11:35 AMtopsquark