Finding a exponent

  • Thread starter thomas49th
  • Start date
  • #1
655
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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

Thankyou
Tom
 

Answers and Replies

  • #2
312
0
square both sides and use logs
 
  • #3
655
0
we dont use log function at GCSE level
 
  • #4
312
0
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)
 
  • #5
655
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how do I use the log function on a calculator? You will need a calculator right?
 
  • #6
123
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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
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
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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

Thankyou
Tom
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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