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LaronJaxon

What is the solution to the equation 1/sqrt 8= 4^(m - 3) ?

  • one year ago
  • one year ago

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  1. zordoloom
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    m=9/4

    • one year ago
  2. terenzreignz
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    \[\large \frac{1}{\sqrt{8}}=4^{m-3}\]

    • one year ago
  3. terenzreignz
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    One nifty way of doing this is to try to express both sides of the equation as exponentials, ie, one base, one exponent... preferably with the same base. So, let's start with \[\frac{1}{\sqrt{8}}\] 8 is just 2³, so let's put it that way...\[\large \frac{1}{\sqrt{2^{3}}}\] And remember that taking the square root means raising something to the 1/2 power, so...\[\huge \frac{1}{\left( 2^3 \right)^{\frac{1}{2}}}\] Using laws of exponents, you get

    • one year ago
  4. terenzreignz
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    \[\huge \frac{1}{2^{\frac{3}{2}}}\]Now remember that \[\large a^{-n}=\frac{1}{a^n}\] So eventually, we're left with \[\huge \frac{1}{\sqrt{8}} = 2^{-\frac{3}{2}}\] Now on to the other side of the equation...

    • one year ago
  5. terenzreignz
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    \[\large 4^{m-3}\] But 4 = 2² So, we can write it as \[\huge (2^2)^{m-3}\] Again, using laws of exponents, it is just equal to \[\huge 2^{2(m-3)}=2^{2m - 6}\]

    • one year ago
  6. terenzreignz
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    So, your problem becomes... \[\huge 2^{-\frac{3}{2}}=2^{2m - 6}\] Which can only mean \[\large -\frac{3}{2}=2m - 6\]

    • one year ago
  7. terenzreignz
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    And the rest, is history :D

    • one year ago
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