Given that log

_{2}x=p and log_{4}y=q, express x^{2}y in terms of p and of q.Ans: 2

^{2p+2q}Thank You :)

Given that log_{2}x=p and log_{4}y=q, express x^{2}y in terms of p and of q.

Ans: 2^{2p+2q}

Thank You :)

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Middletown, CT

Hi;

log_{2}x=p and log_{4}y=q

We have a log to the base of 2.

We have a log to the base of 4.

We would like each to be of the same base.

I choose base of 2. You will understand why below.

log_{4}y=(log_{2}y)/(log_{2}4)

log_{2}4=2 because 2^{2}=4

log_{2}y/2

(1/2)log_{2}y

log_{2}x=p and (1/2)log_{2}y=q

x^{2}y

log ab=log a+log b

log_{2}x^{2}y=log_{2}x^{2}+log_{2}y

log_{2}x^{2}y=2log_{2}x+log_{2}y

Let's multiply both sides by (1/2)

(1/2)log_{2}x^{2}y=log_{2}x+[(1/2)log_{2}y]

(1/2)log_{2}x^{2}y=p+q

Let's multiply both sides by 2...

log_{2}x^{2}y=2p+2q

Let's move the base of 2 over to the right side...

x^{2}y=2^{2p+2q}

Alexandria, VA

The best approach to this problem is to use the fact that the inverse of log_{2}x is exponentiation with 2, and the inverse of log_{4}y is exponentiation with 4. Thus

2^{p} = x and 4^{q} = y .

4 = 2^{2} , so 4^{q} = (2^{2})^{q }
(a power to a power) so 4^{q} = 2^{2q}

From this it is clear that x^{2} y = 2^{2p} 2^{2q} = 2^{2p+2q}

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