base is 2 for the logs but i cant make 3 a base of 2. so how do i solve for x?
2007-09-19 15:23:19 · 7 answers · asked by C.L. ~Raptors~ 4 in Science & Mathematics ➔ Mathematics
the answer to log(2)x + log(2)3 = 3 is 8/3
can someone show me how to get 8/3 as the answer
2007-09-19 15:27:42 · update #1
northstar what do u mean exponentiate? i dont get how u got 2^3 from 3x...
2007-09-19 15:30:06 · update #2
ok i know using the product rule youll get
log(3x)=3
but how do you get 8/3 from that? is there another rule?
2007-09-19 15:34:07 · update #3
OH YEAH YOU GOTTA CONVERT TO EXPONENTIAL FORM FROM THE LOG FORM, dammit how could i forget it, thanks guys
2007-09-19 15:38:39 · update #4
Let log mean log to base 2 in the following:-
log x + log 3 = 3
log (3 x) = 3
2 ³ = 3 x (from definition of logarithm)
3 x = 8
x = 8 / 3
2007-09-19 22:24:39 · answer #1 · answered by Como 7 · 2⤊ 0⤋
Solve for x. All logs are base 2.
log x + log 3 = 3
log(3x) = 3
Exponentiate both sides.
3x = 2³ = 8
x = 8/3
2007-09-19 22:27:54 · answer #2 · answered by Northstar 7 · 2⤊ 0⤋
log(2) x + log(2) 3 = 3
log[2](3x) = 3 (since log a + log b = log ab )
2^3 = 3x (since if log(10) 100 =2, then 10^2 = 100)
3x =8
x = 8/3
2007-09-19 22:34:55 · answer #3 · answered by mohanrao d 7 · 0⤊ 1⤋
log(2)x + log(2)3 =3
which is the same as
log(2)3x=3
[when adding, you always multipy the two, if minus, it's divide]
and then you would multiply 3x by 3 and make it equal to the base, 2, giving you
3x^3=2
divide two by three
x^3=2/3
2^3=8
so x=8/3
=]
2007-09-19 22:36:52 · answer #4 · answered by missesalexturner 1 · 1⤊ 1⤋
log (2) x + log (2) 3 = 3
log (2) x.3 =3
it means that 3 is an exponent, with base 2, in order to obtain x.3:
2^3=x.3 = 3x
8=3x
8/3=x
2007-09-19 22:33:32 · answer #5 · answered by ecosierra51 2 · 0⤊ 1⤋
log(a)+log(b)=log(ab)
log(2)x + log(2)3 = 3
log(2)[3x]=3
3x=2^3=8
x=8/3
2007-09-19 22:30:08 · answer #6 · answered by ptolemy862000 4 · 0⤊ 0⤋
use the Product rulee mann i swear im learning this **** right now grade 12 advanced functios this stuff is killerr!
2007-09-19 22:27:27 · answer #7 · answered by Ritesh P 2 · 0⤊ 2⤋