Logarithms

Question

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

 

 

 

Ans: 2^2p+2q

 

 

Thank You :)

Answer 1

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=2^2p+2q

Answer 2

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

 

2^p =  x   and     4^q  = y   .

   4 =  2²  ,   so    4^q =  (2²)^q      (a power to a power)  so  4^q =  2^2q

 

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