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# Logarithms

Given that log2x=p and log4y=q, express x2y in terms of p and of q.

Ans: 22p+2q

Thank You :)

Arthur;
How did I do?
It looks good to me, Vivian. You used the rule for multiplication and the change of base rule. I never would have thought of that, but the rule for multiplication was fresh in your memory from the last problem. I try to solve the problem the simplest way so I probably would have used the definition of logarithm like Richard P. did. However, there are many ways to solve a problem and it's good for the students to see the problem done in a different way. Many times I do a problem by whatever method comes to mind first and then I see it done differently by someone else.
Arthur

Hi;
log2x=p and log4y=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.
log4y=(log2y)/(log24)
log24=2 because 22=4
log2y/2
(1/2)log2y
log2x=p and (1/2)log2y=q

x2y
log ab=log a+log b
log2x2y=log2x2+log2y
log2x2y=2log2x+log2y
Let's multiply both sides by (1/2)
(1/2)log2x2y=log2x+[(1/2)log2y]
(1/2)log2x2y=p+q
Let's multiply both sides by 2...
log2x2y=2p+2q
Let's move the base of 2 over to the right side...
x2y=22p+2q
The best approach to this problem is to use the fact that the inverse of log2x  is exponentiation with 2, and the inverse of log4y  is exponentiation with 4.  Thus

2p =  x   and     4q  = y   .
4 =  22  ,   so    4q =  (22)q      (a power to a power)  so  4q =  22q

From this it is clear that x2 y  =  22p 22q   =   22p+2q