1/2 log2 x^4 +1/4 log2 x^4 - 1/6 log2 x?

4 ⤊

1/2 log2 x^4 +1/4 log2 x^4 - 1/6 log2 x?

2007-12-02 06:33:35 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

3 answers

log is to base 2 in the following:-
(1/2) log (x^4) + (1/4) log (x^4) - (1/6) log (x)
log x² + log x + log x^(-1/6)
log x³ + log x^(-1/6)
log x ^(17/6)

2007-12-02 06:54:07 · answer #1 · answered by Como 7 · 3⤊ 1⤋

1/2 log2 x^4 +1/4 log2 x^4 - 1/6 log2 x
= log2 x^2 + log2 x^1 - log2 x^1/6
= log2 ( x^2*x^1/x^1/6)
= log2 x^ (2 5/6)

2007-12-02 06:39:45 · answer #2 · answered by harry m 6 · 1⤊ 1⤋

1/2 log2 x^4 +1/4 log2 x^4 - 1/6 log2 x
log2 (x^4)^(1/2) +log2 (x^4)^(1/4) - log2 x^(1/6)
log2 x^2 + log2 x - log2 x^(1/6)
log2 ((x^2) x (x) / (x^(1/6)))
log2 x^(2+1-1/6)
log2 x^(17/6)
= (17/6)log2 x

2007-12-02 06:43:47 · answer #3 · answered by Wits 2 · 0⤊ 0⤋