- #1

- 655

- 0

[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

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- Thread starter thomas49th
- Start date

- #1

- 655

- 0

[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

- #2

- 312

- 0

square both sides and use logs

- #3

- 655

- 0

we dont use log function at GCSE level

- #4

- 312

- 0

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

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- #7

HallsofIvy

Science Advisor

Homework Helper

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You should know that [itex]\sqrt{x}= x^{\frac{1}{2}}[/itex]

[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

Since 128= 2

Your equation [itex]\sqrt{128}= 2^m[/itex] is equivalent to [itex]2^{\frac{7}{2}}= 2^m[/itex] and, then, since 2

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