2006-11-29 19:18:45 · 9 answers · asked by DEBRA S 1 in Science & Mathematics ➔ Mathematics
In the logarithm, there are several number of rules you have to remember. One of them is by combining two or more logarithms into one.
For example,
log (a) b + log (a) c = log (a) bc
log (a) b - log (a) c = log (a) (b/c)
a is the base of the logarithm.
b and c is any value more than 0.
This kind of combination can only be done if and only if the logarithms have the same base.
The other rule involved in solving your question is as following:
c log (a) b = log (a) b^c
a is the base of the logarithm.
b is any value more than 0.
c is any real number.
In your case,
log (a) (y + 5) + 2 log (a) (x + 1)
= log (a) (y + 5) + log (a) (x + 1)^2
= log (a) (y + 5)(x + 1)^2
If you would like to multiply it,
log (a) (y + 5)(x + 1)^2
= log (a) (y + 5)(x^2 + 2x + 1)
= log (a) (x^2y + 2xy + y + 5x^2 + 10x +5)
= log (a) (x^2y + 5x^2 + 2xy + y + 10x +5)
However, this only further complicate the answer, therefore log (a) (y + 5)(x + 1)^2 would be the best answer.
2006-11-29 20:00:06 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Few policies to bear in mind whilst condensing logarithms: a log b = log b^a - skill Rule log b + log a = log ba = log (b*a) - Addition Rule log b - log a = log (b/a) - Subtraction Rule 2 log x + 4 log y - 2 log z skill Rule: log x^2 + log y^4 - log z^2 Addition Rule for the 1st 2 words: log (x^2 * y^2) - log z^2 Subtraction Rule: log ((x^2-y^2)/z^2)
2016-10-04 13:26:39 · answer #2 · answered by elidia 4 · 0⤊ 0⤋
log a (y+5)+2 log a (x+1)
=log a(y+5)+log a (x+1)^2
=loga((y+5)*(x+1)^2)
2006-11-29 19:24:08 · answer #3 · answered by riya s 2 · 0⤊ 0⤋
log a + log b = log ab So,
log a (y+5)(x+1) + 2
log a (xy+y+5x+5) +2
2006-11-29 19:22:34 · answer #4 · answered by Anonymous · 0⤊ 0⤋
2log(a+1)=log(a+1)^2 power rule
log (y+5)+2log(a+1)
=log(y+5)(a+1)^2 product rule
2006-11-29 19:24:06 · answer #5 · answered by raj 7 · 0⤊ 0⤋
Remember
log a.b=loga + log b
Then
log_a ( (y+5). (x+1)^2 )
2006-11-29 19:21:43 · answer #6 · answered by iyiogrenci 6 · 0⤊ 0⤋
loga (y+5) + 2 loga (x+1)
= loga (y+5) + loga (x+1)^2
= loga (y+5)(x+1)^2
= loga (y+5)(x^2 + 2x + 1)
= loga (x^2*y + 2xy + y + 5x^2 + 10x + 5)
2006-11-29 19:26:13 · answer #7 · answered by horensen 4 · 0⤊ 0⤋
log a(y+5)+log a(x+1)^2
log x+ log y = log xy
so, it will be
log a^2(y+5)(x+1)^2
2006-11-29 19:31:09 · answer #8 · answered by fii 3 · 0⤊ 0⤋
loga(y+5)+2loga(x+1)
loga(y+5)+loga^2(x+1)^2
log(a^3(y+5)(x+1)^2)
2006-11-29 19:25:07 · answer #9 · answered by Badri 2 · 0⤊ 0⤋