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Thread: [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: $\displaystyle a^x > 0$


    b) First things first: $\displaystyle x \neq 6$

    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 $\displaystyle x > 6$ and we do not need to worry about the direction of the inequality changing

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

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

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

    Multiply by 4: .$\displaystyle 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: .$\displaystyle \left(2^x - 1\right)\left(2^x - 2\right) \:=\:0$


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

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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 $\displaystyle a^n \cdot a^m = a^{n + m}$.
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