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1. find the value of m when: sqaure root 128 = 2 to the power of m

How in the world do I do this? What are the basic steps please someone tell me
2. (Original post by jasons61)

find the value of m when: sqaure root 128 = 2 to the power of m

How in the world do I do this? What are the basic steps please someone tell me
2^m = 128^0.5

128 can be expressed as a power of 2: 128 = 2^7

2^m = (2^7)^0.5

2^m = 2^3.5 (multiply powers)

Bases are the same, therefore powers are the same => m = 3.5
3. (Original post by jasons61)

find the value of m when: sqaure root 128 = 2 to the power of m

How in the world do I do this? What are the basic steps please someone tell me
Ignore the square root to start with, and concentrate on the 128. Starting with 2, what power of 2 is 128? Well 128 = 2x2x2x2x2x2x2 = 2^7. The square root of a number x = x^(1/2). Can you finish from there?
4. this is nice and easy

(128)^0.5 = 2^m

128 = 2^2m

now there are two methods one is basic other is avanced:

log both sides to get:

log(128) = 2m (log2)

log(128)/log(2) = 2m

7 = 2m

m = 7

method 2 (easy)

(128) = 2^2m

you shoudl realise that 128 also equals 2 to the power of 7

so 2^7 = 2#^2m

therefore: 7 = 2m

and m = 7, as before.

5. I don't know the step by step method... but you can work it out just by observation..

2^m = (128)^0.5

2^m = (2^X)^0.5

where (2^X) = 128
>That calculation is easy...

as 128
64 (2^6)
32 (2^5
16 (2^4
8 (2^3
4 (2^2
2 (2^1

ie 2^7

Therefore 2^m = (2^7)^0.5

Times The Powers

Therefore, 2^m = 2^ 3.5

Therefore m = 3.5
6. i got a better one
ROOT 128 = 2^m
128 = 2^2m
log 128 = 2m log 2
log 128 / log 2 = 2m
7 = 2m
m = 3.5

cut the boring factorisation
7. Looks like he's doing GCSE, so won't need logs..
8. I found a better method, a P1 method. Im gonna use ¬ as squared root.

What you do is simplify the ¬128.
= ¬128 = ¬64 x 2 = 8¬2

Therefore 8¬2 = 2^m

From P1, you should know that a¬b = b^1/a

Therefore using this method m = 1/8

9. Your use of a¬b = b^(1/a) isn't right....
e.g.
b^(1/3) is the cube root of b, not 3x¬b
10. (Original post by another_katie)
Your use of a¬b = b^(1/a) isn't right....
e.g.
b^(1/3) is the cube root of b, not 3x¬b

I never said there was an x involved.
11. sorry, i meant 3 times ¬b
(3 lots of root b) isn't equal to b^(1/3) because that's the cube root of b, not three times the square root of it....

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