Finding Exponent for \sqrt{128} = 2^m

[thomas49th]

Hi, I have the equation


$$\sqrt{128} = 2^{m}$$

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

$$\sqrt{128} = 2^{m}$$

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

Thankyou
Tom

 

[turdferguson]

square both sides and use logs

 

[thomas49th]

we don't use log function at GCSE level

 

[turdferguson]

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)

 

[thomas49th]

how do I use the log function on a calculator? You will need a calculator right?

 

[jing]

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

 

[HallsofIvy]

Hi, I have the equation


$$\sqrt{128} = 2^{m}$$

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

$$\sqrt{128} = 2^{m}$$

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

Thankyou
Tom

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