please show work.
2007-01-17 12:18:49 · 3 answers · asked by huronda_hottie_2006 1 in Science & Mathematics ➔ Mathematics
log[base 4] (2√32) + log[base 27] (√3)
= log[base 4] (8√2) + log[base 27] (√3)
= log[base 4] {2^(7/2)} + log[base 27] {3^(1/2)}
= log[base 4] {4^(7/4)} + log[base 27] {27^(1/6)}
= 7/4 + 1/6 = (21 + 2)/12 = 23/12
2007-01-17 14:00:02 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
log4 (2 sqrt 32) + log27 sqrt 3=log(2^2) (2^(7/2)) + log(3^3) 3^(1/2)=7/4+1/6=(21+2)/12=23/12
2007-01-17 20:29:59 · answer #2 · answered by 121221 2 · 0⤊ 0⤋
log{4} (2) (32)^(1/2) + log{27} (3)^(1/2)
= 2 log{2} (2) (16)^(1/2) (2)^(1/2) + 3 log{3} (3)^(1/2)
= 16 (2)^(1/2) log{2} + 3 (3)^(1/2) log{3}
= 2^(9/2) log{2} + 3^(3/2) log{3}
= 9.290726, if the log is base 10.
= 21.39268, if the log is base e.
2007-01-17 20:42:09 · answer #3 · answered by Anonymous · 0⤊ 0⤋