Results 1 to 3 of 3

Math Help - Solve Logarithm Inequality

  1. #1
    Junior Member
    Joined
    Dec 2008
    Posts
    34

    Solve Logarithm Inequality

    Hi--I have the following problem:

    Solve the inequality log2(112)is less than x. (2 is the base)
    I did this problem and got x is less than 4. However, I was told this is the incorrect answer. It should be x is more than 4. I thought that you only had a change in the equality sign when you are multiplying or dividing by a negative number. Why did this inequality sign change direction?

    Thanks for explaining it to me.

    Joanie
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, Joanie!

    If no one is responding, it's that we don't understand the problem.
    And the answers (yours and theirs) are baffling.

    Could you give us the original wording?


    Solve the inequality: . \log_2(112)\: < \:x . ??
    It says: . x \:> \:\log_2(112) . . . It's already solved!


    By the way, that's how they "changed the sign" .

    If we have: .  2 < 7 .("Two is less than seven")
    . . we can write: . 7 > 2 .("Seven is greater than two")

    We didn't "change the sign", we reversed the entire statement.


    "John is taller than Mary" .= ."Mary is shorter than John." . . . . Get it?


    I did this problem and got: . x < 4 . ??
    How did you do that?


    However, I was told it should be: . x > 4 . ??
    How did they do that?


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Is a puzzlement!


    It turns out that: \log_2(112) \:\approx\:6.81

    So they gave us: . x \:>\:6.81

    And they ask us to "solve for x" . . . Rather silly, isn't it?

    Then they shout, "You're wrong! .The answer is: x > 4"


    Someone has been taking too much cough syrup . . .

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by Joanie View Post
    Hi--I have the following problem:

    Solve the inequality log2(112)is less than x. (2 is the base)
    \log_2112<x? There is nothing here to solve.

    I did this problem and got x is less than 4. However, I was told this is the incorrect answer. It should be x is more than 4.
    Assuming you gave the correct inequality, x>4 does not work. For example, 2^5=32<112, so x=5 does not satisfy the inequality. x<4 similarly does not work.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inequality with logarithm, help needed
    Posted in the Algebra Forum
    Replies: 4
    Last Post: August 15th 2010, 05:15 AM
  2. Solve the logarithm.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 22nd 2010, 01:12 PM
  3. Solving inequality with logarithm
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 11th 2009, 02:59 PM
  4. Inequality with logarithm help needed.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 24th 2009, 04:15 AM
  5. Solve for x logarithm
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 17th 2008, 07:10 AM

Search Tags


/mathhelpforum @mathhelpforum