# solution set of the logarithm inequality

• Dec 22nd 2012, 06:20 AM
rcs
solution set of the logarithm inequality
Solution set: Log base 1/3 (x^2+ 4x +3) < = -1

attempt:
x^2 + 4x + 3 <= (1/3)^-1
x^2 + 4x + 3 < = 3
x^2 + 4x < = 0
x ( x + 4) < = 0
x < = 0, x < = -4

i wonder why from the book i have read, the solution set is I.N. (-4, 0) or - 4 < x < 0

please guide me on this problem

thanks
• Dec 22nd 2012, 08:35 AM
BobP
Re: solution set of the logarithm inequality
At the point where you have x(x+4)<=0, sketch the graph of y = x(x+4).
The range of values for which y<=0 should be obvious.
• Dec 22nd 2012, 08:37 AM
ILikeSerena
Re: solution set of the logarithm inequality
Hi rcs!

Quote:

Originally Posted by rcs
Solution set: Log base 1/3 (x^2+ 4x +3) < = -1

attempt:
x^2 + 4x + 3 <= (1/3)^-1

This is incorrect.
When the base of the log is between 0 and 1 the inequality swaps around.

Quote:

x^2 + 4x + 3 < = 3
x^2 + 4x < = 0
x ( x + 4) < = 0
x < = 0, x < = -4
The last step is incorrect.
If x would be smaller than -4, say -5, then both factors are negative, resulting in a positive number.
Note that a product is negative, if and only if the factors have opposite signs.

Quote:

i wonder why from the book i have read, the solution set is I.N. (-4, 0) or - 4 < x < 0

please guide me on this problem

thanks
Btw, are you aware that you cannot take the log of zero or a negative number?
That is, you have the additional constraint that x^2+ 4x +3 > 0.
It means that when you have solved your inequality, you need to verify that this constraint is satisfied.
• Dec 22nd 2012, 08:51 AM
rcs
Re: solution set of the logarithm inequality
Quote:

Originally Posted by ILikeSerena
Hi rcs!

This is incorrect.
When the base of the log is between 0 and 1 the inequality swaps around.

The last step is incorrect.
If x would be smaller than -4, say -5, then both factors are negative, resulting in a positive number.
Note that a product is negative, if and only if the factors have opposite signs.

Btw, are you aware that you cannot take the log of zero or a negative number?
That is, you have the additional constraint that x^2+ 4x +3 > 0.
It means that when you have solved your inequality, you need to verify that this constraint is satisfied.

so if you were to solve it

how would it be done?

can you show it to me please...
• Dec 22nd 2012, 08:56 AM
ILikeSerena
Re: solution set of the logarithm inequality
Ah, well, I don't like giving full solutions.
But perhaps you can redo the calculation taking my remarks into account?
• Dec 22nd 2012, 07:18 PM
rcs
Re: solution set of the logarithm inequality
Quote:

Originally Posted by ILikeSerena
Ah, well, I don't like giving full solutions.
But perhaps you can redo the calculation taking my remarks into account?

if i would redo it how does the equation look like?
• Dec 22nd 2012, 07:29 PM
rcs
Re: solution set of the logarithm inequality
i just wonder why the symbol of inequality from <= changed to < only? how come?
• Dec 23rd 2012, 04:22 AM
ILikeSerena
Re: solution set of the logarithm inequality
Quote:

Originally Posted by rcs
i just wonder why the symbol of inequality from <= changed to < only? how come?

In this particular case it did not change from <= to <.
What you give as a solution is incorrect.
The proper solution is the complement of the interval (-4,0).
That is: x <= -4 or x >= 0.