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Math Help - negative fractional exponents and logarithms

  1. #1
    Member MathBlaster47's Avatar
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    negative fractional exponents and logarithms

    I have a two questions I need a hand with, both of which I think I have a handle on but, as usual, would like to make sure of.

    1:
    Solve the equation.
    2x^{-\frac{1}{2}}=8
    Working:
    2x^{-\frac{1}{2}}=8\rightarrow 2 \frac{1}{x^{\frac{1}{2}}}=8 \rightarrow 2\frac{1}{\sqrt{x}}=8 \rightarrow \frac{2}{2\sqrt{x}}=8.

    I am at something of a loss as to how to finish. Logically speaking, I think that 64 is the answer because 2\cdot\sqrt{64}=16 so \frac{2}{2\cdot\sqrt{64}} should equal 8.
    Better safe than sorry, I think.

    2:
    Express as a logarithm of a single number or expression.
    3\log_{10} 6

    Working(?)
    I think all I have to do is multiply \log_{10} 6 by three right?

    Thank you!
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  2. #2
    Junior Member StonerPenguin's Avatar
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    Hmm, I'll explain later (and I might be wrong) but my answer were;
    2x^{-\frac{1}{2}}=8 and x=\frac{1}{16}
    3\log_{10} 6 = log_{10} 216
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by MathBlaster47 View Post
    I have a two questions I need a hand with, both of which I think I have a handle on but, as usual, would like to make sure of.

    1:
    Solve the equation.
    2x^{-\frac{1}{2}}=8
    Working:
    2x^{-\frac{1}{2}}=8\rightarrow 2 \frac{1}{x^{\frac{1}{2}}}=8 \rightarrow 2\frac{1}{\sqrt{x}}=8 \rightarrow \frac{2}{2\sqrt{x}}=8.

    I am at something of a loss as to how to finish. Logically speaking, I think that 64 is the answer because 2\cdot\sqrt{64}=16 so \frac{2}{2\cdot\sqrt{64}} should equal 8.
    Better safe than sorry, I think.

    2:
    Express as a logarithm of a single number or expression.
    3\log_{10} 6

    Working(?)
    I think all I have to do is multiply \log_{10} 6 by three right?

    Thank you!
    1. a^{-\frac{m}{n}} = \frac{1}{a^{m/n}} = a^{n/m}

    2x^{-1/2} = \frac{2}{x^{1/2}} = \frac{2}{\sqrt{x}}

    I'm not sure where the 2 on the denominator comes from

    \frac{2}{\sqrt{x}} = 8 then \frac{1}{\sqrt{x}} = 4

    Take the reciprocal of both sides

    \sqrt{x} = \frac{1}{4}

    x = \frac{1}{16}


    ======================================


    2. Use the log power law:

    a\, \log _b(c) = \log _b (c^a)

    3\log_{10}(6) = \log_{10} (6^3) = \log_{10} (216)
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  4. #4
    Member MathBlaster47's Avatar
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    Quote Originally Posted by StonerPenguin View Post
    Hmm, I'll explain later (and I might be wrong) but my answer were;
    2x^{-\frac{1}{2}}=8 and x=\frac{1}{16}
    3\log_{10} 6 = log_{10} 216
    Your answer is corroborated by e^(i*pi), good job and thanks!

    Quote Originally Posted by e^(i*pi) View Post
    1. a^{-\frac{m}{n}} = \frac{1}{a^{m/n}} = a^{n/m}

    2x^{-1/2} = \frac{2}{x^{1/2}} = \frac{2}{\sqrt{x}}

    I'm not sure where the 2 on the denominator comes from

    \frac{2}{\sqrt{x}} = 8 then \frac{1}{\sqrt{x}} = 4

    Take the reciprocal of both sides

    \sqrt{x} = \frac{1}{4}

    x = \frac{1}{16}


    ======================================


    2. Use the log power law:

    a\, \log _b(c) = \log _b (c^a)

    3\log_{10}(6) = \log_{10} (6^3) = \log_{10} (216)
    For 1 I got a little mixed up, and multiplied both sides by 2, so I clearly need to slow down when working my problems out....
    As for 2.....ah........making note of my errors.....rereading text book......giving self double facepalm in mirror......ok I understand what happened!
    Thank you again guys!
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  5. #5
    Junior Member StonerPenguin's Avatar
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    Quote Originally Posted by MathBlaster47 View Post
    For 1 I got a little mixed up, and multiplied both sides by 2, so I clearly need to slow down when working my problems out....
    As for 2.....ah........making note of my errors.....rereading text book......giving self double facepalm in mirror......ok I understand what happened!
    Thank you again guys!
    Ah, don't be so hard on yourself, I did something similar with the question I recently asked on here, I was stupid and made things way harder than they needed to be. Lawl DUUURRHUUURR When you're homeschooled it's good have 'live bodies' to ask about this kinda stuff. It's all good, 'cuz you're learning, amirite?
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  6. #6
    Member MathBlaster47's Avatar
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    No worries! Just laughing off my mistakes!
    Live and learn, mostly to slow down and double check my working!
    Yep, is really good to have live bodies to bounce stuff off of!
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