solve : log x^2 to the base(2x+3) is less than 1
asked Sep 19, 2017 in Algebra 1 Answers by Paula

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

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, 2017 by Rod Top Rated User (552,540 points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
81,208 questions
85,305 answers
2,154 comments
68,736 users