Express 2 log 4 – 3 log 5 + 4 log 6 as a single logarithm.
a) log 12.8
b) log (4.9 x 10^-5)
c) log 1187
d) log 165.888
NOTE: There are no bases in this question.
2007-03-08 01:53:15 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
log4^2-log5^3+log6^4
log 16-log125+log1296
log16/125+log1296
log16*1296/125= log 165.888
youjust use the power of log and product and quotient of log
loga^n=nloga
loga+logb=loga*b
loga-logb=loga/b
good luck
2007-03-08 01:59:47 · answer #1 · answered by Helper 6 · 0⤊ 0⤋
2 log 4 - 3 log 5 + 4 log 6
=log4^2 -log5^3+log6^4
=log(4^2/5^3*6^4)
=log 165.888
d.
2007-03-08 02:02:02 · answer #2 · answered by pigley 4 · 0⤊ 0⤋
2log[9] (something) = log[3] (the comparable element). This comes from the definition of logarithm and the actuality that 3^2 = 9. So multiply the two facets of your equation via 2 and get: log[3](6x^2 - 19x + 2) = 2log[3](2-3x) or, using a property of logs: log[3](6x^2 - 19x + 2) = log[3](2-3x)^2, so 6x^2 - 19x + 2 = (2-3x)^2, now remedy this equation, and verify to envision strategies because of the fact strategies (if any) that make the two (2 - 3x) or (6x^2 - 19x + 2) damaging ought to be thrown out, because of the fact you could no longer take logs of damaging numbers.
2016-12-18 17:58:47 · answer #3 · answered by ? 4 · 0⤊ 0⤋
it's d) log 165.888
it's log(4^2) - log(5^3) + log(6^4)
= log (16*1296/125)=d)
2007-03-08 01:58:51 · answer #4 · answered by A New Life 3 · 0⤊ 0⤋
The answer is d.
a log(b) = log (b^a)
log a + log b = log (axb)
log a - log b = log (a/b)
2007-03-08 02:07:48 · answer #5 · answered by Anonymous · 0⤊ 0⤋
d)
2007-03-08 01:58:16 · answer #6 · answered by Biz Iz 3 · 0⤊ 0⤋