Prove this, using laws of exponents?

Question

[4^(m+1/4) * {2^(m+1)}1/2 over 2 * 2^(-m/2)] 1/m = 8

Answer 1

4^(m+1/4) * 2^{(m+1)/2}
= 2^(2m+1/2) * 2^{(m+1)/2}
= 2^{(2m + 1/2) + 0.5(m + 1)}
= 2^(1.5m + 1)

2 * 2^(-m/2)
= 2^(1 - 0.5m)

Now,
[4^(m+1/4) * 2^{(m+1)/2}] / [2 * 2^(-0.5m)]
= {2^(1.5m + 1)} / {2^(1 - 0.5m)}
= 2^{(1.5m + 1) - (1 - 0.5m)}
= 2^(3m)

So LHS
= [{4^(m+1/4) * 2^((m+1)/2)} / {2 * 2^(-0.5m)}]^(1/m)
= [2^3m]^(1/m)
= 2^3
= 8
= RHS

Hence proved.

Answer 2

You need to be extremely careful in complicated expressions like this to get all your parentheses, etc. in place so the expression is completely unambiguous. In this case, you left out a couple exponent (^) signs and some parens. The expression should be:

[(4^(m+1/4) * {2^(m+1)}^(1/2))/(2 * 2^(-m/2))]^ (1/m) = 8

The rules for evaluation are to do the exponents, then multiply-divide, then add-subtract. Any expression in parenheisis must be completed by itself.

Begin with: {2^(m+1)}^(1/2). You know that multiple exponents simpy multiply, that is: (A^B)^C = A^(B*C) so:

{2^(m+1)}^(1/2)) = 2^(m/2 + 1/2)

Next look at: 4^(m+1/4). Since 4 = 2^2 you can rewrite this as (2^2)^(m + 1/4)

4^(m+1/4) = (2^2)^(m + 1/4)

Again the the multiple exponents multiply so:

4^(m+1/4) = 2^(2*m + 1/2)

Now combine these two terms by repolacing them with the equivalents just derived:

4^(m+1/4) * {2^(m+1)}^(1/2) = (2^(2*m + 1/2))*(2^(m/2 + 1/2))

The right hand side is all in powers of 2 now. I rewrote 4 as 2^2 to do that. I can combine the exponents since adding exponents is the same as multiplying the terms:
A^(B + C) = A^B + A^C so:

(2^(2*m + 1/2))*(2^(m/2 + 1/2)) = 2^((2*m + 1/2)+(m/2 + 1/2)) = 2^((2 + 1/2)*m + 1)

Now on to the denominator: 2 * 2^(-m/2)

Since 2 = 2^1 and multiplying terms adds the exponents:

2 * 2^(-m/2) = 2^(1 - m/2)

To combine this with the numerator realize that to bring the numberator to the top negates the exponent. That is: 1/(A^B) = A^(-B) so:

1/(2 * 2^(-m/2)) = 1/(2^(1 - m/2)) = 2^(-(1 - m/2)) = 2^(m/2 - 1)

All that's left now is to multiply the to combined terms and do the final exponent. Combnining everything so far:

(2^((2 + 1/2)*m + 1) * 2^(m/2 - 1))^(1/m)

Begin by adding exponents:

((2 + 1/2)*m + 1) + (m/2 - 1) = 3*m

So it reduces to:

(2^(3*m))^(1/m)

Multiply the multiple exponents:

(2^(3*m))^(1/m) = 2^((3*m) * (1/m)) = 2^3 = 8

Answer 3

[4^(m+1/4) * {2^(m+1)}1/2 over 2 * 2^(-m/2)] 1/m, as written, is not equal to 8. The entire expression inside square brackets will be equal to 2 raised to some power. When you multiply it by (1/m), you get some power of 2 over m, which will definitely not be reducible to 8. Perhaps there is a typo in your question.

If you reformat the question as written to instead read [(4^(m + 1/4)*2^((m + 1)/2)) / (2*2^(-m/2))]^(1/m), then the solution by the previous answerer applies.