log4a+log4b
log3m-log3n
3log5 2x
log3r+log3s+log3t
3log2r-2log2s
1/2(log10x-log10y)
2007-12-19 09:57:55 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
log (4a * 4b) -- then simplify what's inside to get log (16ab)
log (3m/3n) -- then simplify
log base 5 ((2x)^3) -- then simplify what's inside to 8x^3
and so on.
On the last one, recall that exponents don't need to be integers. Raising something to the 1/2 power is just taking its square root. (The "something" will be x/y)
2007-12-19 13:09:51 · answer #1 · answered by Curt Monash 7 · 0⤊ 0⤋
log { [x^(5/6) * y^(2/3)] / [x^(a million/2) * y]} Then in basic terms the indices: log { x^(a million/38c5dee9189b26f774614ce1d6ae711^(a million/3)} it is truly useful to take the a million/3 out of the log log (x/y) / 3 i think of you have got completed that, have self belief - believe the algebra.
2016-12-11 09:32:44 · answer #2 · answered by carmean 4 · 0⤊ 0⤋