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Math Help - [SOLVED] Checking Inequalities

  1. #1
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    [SOLVED] Checking Inequalities

    Sorry I'm quite bad at maths, but wanted to find out

    a) 4^(x - 1) + 0.5 = (3/4).2^x

    and

    b) 4/(x - 6) > 5

    thanks
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  2. #2
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    Quote Originally Posted by anindyog View Post
    Sorry I'm quite bad at maths, but wanted to find out

    a) 4^(x - 1) + 0.5 = (3/4).2^x

    and

    b) 4/(x - 6) > 5

    thanks
    The first one is an equation, not an inequality. Remember that for real x and positive a: a^x > 0


    b) First things first: x \neq 6

    Since the right hand side is positive then it follows that the left hand side must also be positive, this sets a lower bound of x > 6 and we do not need to worry about the direction of the inequality changing

    This gives 4 > 5x-30 and you can solve for x in the normal way.

    x < \frac{34}{5}
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  3. #3
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    Hello, anindyog!

    (a)\;\; 4^{x - 1} + \tfrac{1}{2}\;=\;\tfrac{3}{4}\!\cdot\!2^x

    Multiply by 4: . 4^x + 2 \:=\:3\!\cdot\!2^x \quad\Rightarrow\quad (2^2)^x + 2 \:=\:3\!\cdot\!2^x \quad\Rightarrow\quad 2^{2x} - 3\!\cdot\!2^x + 2 \:=\:0

    Factor: . \left(2^x - 1\right)\left(2^x - 2\right) \:=\:0


    Therefore: . \begin{Bmatrix}\;2^x-1\:=\:0 & \Rightarrow & 2^x \:=\:1 & \Rightarrow & x \:=\:0\; \\ \\[-3mm]2^x-2 \:=\:0 & \Rightarrow & 2^x \:=\:2 & \Rightarrow & x \:=\:1 \end{Bmatrix}

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  4. #4
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    Thanks alot! However how do you get to 4^x from 4^(x-1) after multiplying it by 4?
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  5. #5
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    Quote Originally Posted by anindyog View Post
    Thanks alot! However how do you get to 4^x from 4^(x-1) after multiplying it by 4?
    You're expected to be familiar with index laws such as a^n \cdot a^m = a^{n + m}.
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