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Thread: Logarithmic inequality

  1. #1
    Member rowe's Avatar
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    Logarithmic inequality

    Solve $\displaystyle x+3^x < 4$ for x.

    Is there an algebraic solution for this? I plugged in values for $\displaystyle x+3^x-4<0$, giving me what I suspect is right, x < 1. But how to be sure?
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  2. #2
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    Quote Originally Posted by rowe View Post
    Solve $\displaystyle x+3^x < 4$ for x.

    Is there an algebraic solution for this? I plugged in values for $\displaystyle x+3^x-4<0$, giving me what I suspect is right, x < 1. But how to be sure?
    1. Graph the function $\displaystyle f(x)=x+3^x-4$ and check for which values of x the values of f(x) are negative.

    2. attempt: Solve
    $\displaystyle x+3^x=4$ for x. Since $\displaystyle y = x \text{ and } y = 3^x$ are monotically increasing there exists only one solution. By first inspection x = 1.
    Therefore the inequality is true for all x < 1.
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  3. #3
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    Quote Originally Posted by rowe View Post
    Solve $\displaystyle x+3^x < 4$ for x.

    Is there an algebraic solution for this? I plugged in values for $\displaystyle x+3^x-4<0$, giving me what I suspect is right, x < 1. But how to be sure?
    Alternatively,

    $\displaystyle x+3^x\ <\ 4$

    $\displaystyle 3^x\ <\ 4-x$

    $\displaystyle f(x)=4-x$ is a line decreasing at $\displaystyle 45^o,$ passing through
    the y-axis at (0,4) and the x-axis at (4,0).

    $\displaystyle g(x)=3^x$ is an increasing function passing through the y axis at (0,1), so it cuts f(x) between x=0 and x=4.

    When x=1, both f(x) and g(x) are 3.

    $\displaystyle g(x)\ <\ f(x)\ for\ x\ <\ 1$
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  4. #4
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    Quote Originally Posted by earboth View Post
    1. Graph the function $\displaystyle f(x)=x+3^x-4$ and check for which values of x the values of f(x) are negative.

    2. attempt: Solve
    $\displaystyle x+3^x=4$ for x. Since $\displaystyle y = x \text{ and } y = 3^x$ are monotically increasing there exists only one solution. By first inspection x = 1.
    Therefore the inequality is true for all x < 1.
    @earboth

    1. it would be much easier if he draw 3^x and 4-x and find values where 4-x is "bigger" then 3^x


    2. I do not what made you think that if they're "monotically increasing there exists only one solution"?

    example: x^3+5 and 3^x, both monotonically increasing but have 2 intersections.
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  5. #5
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    Hello, rowe!

    Solve: .$\displaystyle x+3^x \:<\: 4$
    There is no algebraic solution, but there is a graphical solution.

    We have: . $\displaystyle 3^x \:<\:4-x$

    The question becomes: when is $\displaystyle y = 3^x$ below $\displaystyle y = 4-x$ ?


    The graphs look like this:
    Code:
                  |
                * |     *
                  *
                  | *  *
                  |   o 
                  | * : *
                  *   :   *
              *   |   :     *
        *         |   :       *
      - - - - - - + - + - - - - * - -
                  |   1           *
                  |

    As you pointed out, the curves intersect at $\displaystyle x = 1.$

    You answer is correct: . $\displaystyle x \:<\:1$

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