If (log2x)2−(log2y)2=log2256 and log2x2−log2y2=log216. Then the value of log2x6y−2 is....
Explanation
Use the substitution log2x=a and log2y=b. Then transform the logarithm equation into
log2x2−log2y2=log216
log22x−log22y=log224
2a−2b=log422
Convert to exponential form
2(a−b)=4
a−b=2...(1)
a=b+2
Substitute equation (1)
(log2x)2−(log2y)2=log2256
(a)2−(b)2=log228
(a+b)(a−b)=log822
(a+b)(a−b)=log822
(a+b)(2)=8
a+b=4
(b+2)+b=4
2b=2
b=1
Therefore the result from equation (1) is a=b+2=1+2=3.
Determine the value of log2x6y−2
log2x6y−2=log2x6+log2y−2
log2x6y−2=6(log2x)+(−2)log2y
log2x6y−2=6a−2b
log2x6y−2=6(3)−2=16