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Math Help - Exponents

  1. #1
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    Question Exponents

    If a^1^/^x = b^1^/^y = c^1^/^z & b^2 = ac, how do I find the value of (x + z)/2y?

    Thanks,

    Ron
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  2. #2
    Senior Member BAdhi's Avatar
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    Quote Originally Posted by rn5a View Post
    If a^1^/^x = b^1^/^y = c^1^/^z
    take logs,

    \frac{\lg a}{x}=\frac{\lg b}{y}=\frac{\lg c}{z}

    substitute x,z with y
    Last edited by BAdhi; February 15th 2011 at 06:23 PM. Reason: added latex tags
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  3. #3
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    Quote Originally Posted by rn5a View Post
    If a^1^/^x = b^1^/^y = c^1^/^z & b^2 = ac, how do I find the value of (x + z)/2y?

    Thanks,

    Ron
    Apply \ln= neperian logarithm: \displaystyle{a^1^/^x = b^1^/^y = c^1^/^z\Longrightarrow \frac{1}{x}\ln a=\frac{1}{y}\ln b=\frac{1}{z}\ln c , and etc.

    Tonio
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  4. #4
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    I don't have any idea about logs. How do I do it without using logs?

    Thanks,

    Ron
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  5. #5
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    Hello, Ron!

    Without logs? . . . This takes some gymnastics.


    \text{If }a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}}\:\text{ and }\:b^2 = ac,\:\text{find the value of: }\:\dfrac{x + z}{2y}

    \text{We have: }\:b^{\frac{1}{y}} \:=\:a^{\frac{1}{x}}
    \text{Raise both sides to the power }y\!:\;(b^{\frac{1}{y}})^y \;=\;(a^{\frac{1}{x}})^y \quad\Rightarrow\quad b \;=\;a^{\frac{y}{x}} .[1]

    \text{We have: }\:b^{\frac{1}{y}} \:=\:c^{\frac{1}{z}}
    \text{Raise both sides to the power }y\!:\;(b^{\frac{1}{y}})^y \;=\;(c^{\frac{1}{z}})^y \quad\Rightarrow\quad b \;=\;c^{\frac{y}{z}} .[2]

    Multiply [1] and [2]: . b^2 \;=\;a^{\frac{y}{x}}\!\cdot\!c^{\frac{z}{x}}


    We are told that: . b^2 \,=\,a\!\cdot\!c

    \text{Hence: }\;a^{\frac{y}{x}}\!\cdot\!c^{\frac{z}{x}} \:=\:a^1\!\cdot\!c^1 \quad\Rightarrow\quad \begin{Bmatrix}\frac{y}{x} \:=\:1 \\ \\[-4mm] \frac{z}{x} \:=\:1 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix} y \:=\:x \\ z\:=\:x \end{Bmatrix}

    . . And we have: . x = y = z


    \displaystyle \text{Therefore: }\:\frac{x+z}{2y} \;=\;\frac{x+x}{2x} \;=\;\frac{2x}{2x} \;=\;1

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