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3 Answers
- hayharbrLv 75 years ago
If log4 (9) = x, then 4^x = 9, right? That's (2^2)^x = 9, or 2 ^ (2x) = 9
Now do log base 2 of both sides
2x log2 (2) = log2 (9)
Log2 2 = 1 so you have 2x = log2 (9), or x = log2 (9) / 2
Sub that back for log4 (9)
log2 (6) - log2 (9) / 2
= log2 (6) - 1/2 log2 (9)
= log2 (6 ÷ √9)
= log2 (6÷3)
= log2 2
= 1
- ?Lv 75 years ago
a = log_2 (6) then 2^a = 6
a = log 6/log 2
b = lob_4 (9) the (2^2)^b = 9
b = log 9/[2log 2]
a-b = [2log 6-log 9]/[2log 2]
= log (36/9)/log 4
= 1
- SqdancefanLv 75 years ago
log2(6) - log4(9)
= (log2(2) + log2(3)) - log2(9)/log2(4)
= 1 + log2(3) - 2log2(3)/(2log2(2))
= 1 + log2(3) - log2(3)
= 1
