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Finding a exponent

  1. Feb 14, 2007 #1
    Hi, I have the equation

    [tex]\sqrt{128} = 2^{m}[/tex]

    I know that m is [tex]\frac{7}{2}[/tex] as 2^7 = 128 from binary.
    However, say the equation was

    [tex]\sqrt{128} = 2^{m}[/tex]

    how do I go about finding m. Can someone show me the technique

  2. jcsd
  3. Feb 14, 2007 #2
    square both sides and use logs
  4. Feb 14, 2007 #3
    we dont use log function at GCSE level
  5. Feb 14, 2007 #4
    Well informally, after squaring both sides you have 128=2^(2m). But youve already identified that 128=2^7. So 7=2m

    What youre doing is finding the log base 2 of both sides. Log base 2 of 128 = 7 because 2^7=128. Log base 2 of 2^(2m)=2m because 2^(2m)=well... 2^(2m)
  6. Feb 15, 2007 #5
    how do I use the log function on a calculator? You will need a calculator right?
  7. Feb 15, 2007 #6
    thomas49th I teach GCSE and you are not expected to know about logs. You are expected to either know that 2^7=128 or to be able to work out the value using what you do know. So if you know 2^3=8, then you multiply by 2 to get 2^4=16 and to keep going until you get 128
  8. Feb 15, 2007 #7


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    You should know that [itex]\sqrt{x}= x^{\frac{1}{2}}[/itex]
    Since 128= 27, then [itex]\sqrt{128}= 2^{\frac{7}{2}}[/itex]
    Your equation [itex]\sqrt{128}= 2^m[/itex] is equivalent to [itex]2^{\frac{7}{2}}= 2^m[/itex] and, then, since 2x is a one-to-one function, we must have m= 7/2.
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