solve : log x^2 to the base(2x+3) is less than 1
asked Sep 19 in Algebra 1 Answers by Paula

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1 Answer

Make each side of the equation an exponent of (2x+3):

x^2<(2x+3)^1 is the same as x^2<2x+3 or x^2-2x-3<0.

(x-3)(x+1)<0. Therefore x-3<0 and x+1>0; or x-3>0 and x+1<0.

The first case is x<3 and x>-1 so -1<x<3 (x is between -1 and 3).

The second case makes x>3 and x<-1, which is impossible.

The solution is -1<x<3. But we have to eliminate x=0 because we can't have log0. Therefore the answer is:

-1<x<0 or 0<x<3. What about the base? Base=2x+3 so 1<base<3 or 3<base<9.

CHECK

Try base=2, so 2x+3=2, x=-1/2: log(x^2)=log(1/4)=log(2^-2)=-2 which is less than 1. So the inequality holds true.

Try base=5, so 2x+3=5, x=1: log(x^2)=log(1)=0 which is less than 1. So the inequality holds.

Try base=9, so 2x+3=9, x=3. We would not expect the inequality to hold because 9 is out of range. Log(x^2)=log(9)=1, which is not less than 1, and the inequality fails as expected.

answered Sep 19 by Rod Top Rated User (486,860 points)
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