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the top expression can be rewritten as, q^p = 16, so q = 16^(1/p)
into the bottom...
log_(16^(1/p))_ 8p = x , yeah, crap notation
16^(1/p)^x = 8p
16^(x/p) = 8p
(x/p)log16 = log8 + log p
x = (p/log16)(log8 + logp)
maybe.
oh ****. i don't think i've answered a single question correctly in like a week. this time, i can't tell the difference between a p and a q.
into the bottom...
log_(16^(1/p))_ 8p = x , yeah, crap notation
16^(1/p)^x = 8p
16^(x/p) = 8p
(x/p)log16 = log8 + log p
x = (p/log16)(log8 + logp)
maybe.
oh ****. i don't think i've answered a single question correctly in like a week. this time, i can't tell the difference between a p and a q.
arktos
p = logq 16 = logq (2)^4 = 4logq 2
p/4 = logq (2)
logq (8q) = logq q + logq 8 = 1 + logq (2)^3 = 1 + 3logq (2)
since logq (2) = p/4
3logq (2) = 3/4 P
1+3logq (2) = 3/4 p + 1
prob wrong, havn't done logs in a while
p/4 = logq (2)
logq (8q) = logq q + logq 8 = 1 + logq (2)^3 = 1 + 3logq (2)
since logq (2) = p/4
3logq (2) = 3/4 P
1+3logq (2) = 3/4 p + 1
prob wrong, havn't done logs in a while
that;s right, thanks
done a few more questions and now im stuck on :
calculate the value of y for which 2log3 y - log3 (y + 4) = 2
i beleive you use the divide law but it went wrong.
i got log (2y/y+4) = 2
think thats the wrong road though, and if its right dunno what to do next
we always get these mixed excersises on topics but this ones far harder than the other log stuff, every Q seems to have me stumped.
im onto Q9 but i could do 6, 7, 8 or 4c
Q6, 7, 8
6) solve giving your answers as fractions the simulataneous equations:
8^y=4^(2x + 3)
log2 y = log2 x +4
7) Find the values of x for which log3 x - 2 log x 3 = 1
8) solve equation
log3 (2-3x) = log9 (6x^2 - 19x + 2)
theres so much log stuff goin on in my head too ive forgotten the basics!
im onto Q9 but i could do 6, 7, 8 or 4c
Q6, 7, 8
6) solve giving your answers as fractions the simulataneous equations:
8^y=4^(2x + 3)
log2 y = log2 x +4
7) Find the values of x for which log3 x - 2 log x 3 = 1
8) solve equation
log3 (2-3x) = log9 (6x^2 - 19x + 2)
theres so much log stuff goin on in my head too ive forgotten the basics!
8) solve equation
log3 (2-3x) = log9 (6x^2 - 19x + 2)
im gonna try on this one
log3 (2-3x) = log3 (6x^2-19x+2)^1/2
log3 (2-3x)^2 = log3 (6x^2-19x+2)
4-12x+9x^2 = 6X^2-19x+2
3x^2+7x-2=0 (remember this is all to the log3)
x^2+7/3x-2/3=0
(x-7/6)^2 - 49/36 - 24/36
x=7/6+-(63/36)^1/2
dont forget the positive rep
log3 (2-3x) = log9 (6x^2 - 19x + 2)
im gonna try on this one
log3 (2-3x) = log3 (6x^2-19x+2)^1/2
log3 (2-3x)^2 = log3 (6x^2-19x+2)
4-12x+9x^2 = 6X^2-19x+2
3x^2+7x-2=0 (remember this is all to the log3)
x^2+7/3x-2/3=0
(x-7/6)^2 - 49/36 - 24/36
x=7/6+-(63/36)^1/2
dont forget the positive rep
6) solve giving your answers as fractions the simulataneous equations:
8^y=4^(2x + 3)
log2 y = log2 x +4
8^y=4^(2x + 3)
log2 y = log2 x +4
(2^3)^y=(2^2)^(2x+3)
therefore 3y=4x+6 y=4/3X+2
log2 (4/3X+2) - log2 x = 4
log2 (4/3X+2)/x = 4
make this ((4/3X+2)/x)= Z for the moment to get rid of the log on the left
log2 Z = 4
using what we know about logs the answer will be 16
so (4/3X+2)/x =16
(4/3X+2) =16x
2=14(2/3)x divide 2 by 14 and 2/3 and you get x
x=3/22
then put x back into equation and you get Y
massimo
8) solve equation
log3 (2-3x) = log9 (6x^2 - 19x + 2)
im gonna try on this one
log3 (2-3x) = log3 (6x^2-19x+2)^1/2
log3 (2-3x)^2 = log3 (6x^2-19x+2)
4-12x+9x^2 = 6X^2-19x+2
3x^2+7x-2=0 (remember this is all to the log3)
x^2+7/3x-2/3=0
(x-7/6)^2 - 49/36 - 24/36
x=7/6+-(63/36)^1/2
dont forget the positive rep
log3 (2-3x) = log9 (6x^2 - 19x + 2)
im gonna try on this one
log3 (2-3x) = log3 (6x^2-19x+2)^1/2
log3 (2-3x)^2 = log3 (6x^2-19x+2)
4-12x+9x^2 = 6X^2-19x+2
3x^2+7x-2=0 (remember this is all to the log3)
x^2+7/3x-2/3=0
(x-7/6)^2 - 49/36 - 24/36
x=7/6+-(63/36)^1/2
dont forget the positive rep
how does it become ^1/2 then ^2 on the left at the start?
log3 (2-3x) = log9 (6x^2-19x+2)
log3 (2-3x) = log3 (6x^2 -19x+2) / log3 (9) changing bases
log3 (2-3x) x log3 (9) = log3 (6x^2 -19x +2) take denominator to other side
log 3 (11-3x) = log3 (6x^2 -19x + 2 ) multiplication rule
11-3x = 6x^2 -19x +2 same base so equal
6x^2 -16x-9 = 0
do the rest, above is prob wrong.
log3 (2-3x) = log3 (6x^2 -19x+2) / log3 (9) changing bases
log3 (2-3x) x log3 (9) = log3 (6x^2 -19x +2) take denominator to other side
log 3 (11-3x) = log3 (6x^2 -19x + 2 ) multiplication rule
11-3x = 6x^2 -19x +2 same base so equal
6x^2 -16x-9 = 0
do the rest, above is prob wrong.
3y=4x+6------(i) as calculated by massimo
log2 y = log2 x +4
log2 y = log2 x +4 log2 2
log2 y = log2 x + log2 2^4
log2 y = log2 x + log2 16
log2 y = log2 16x
y=16x-------(ii)
solving simultneously we get x=3/22, y=24/11(improper fraction forn).
log2 y = log2 x +4
log2 y = log2 x +4 log2 2
log2 y = log2 x + log2 2^4
log2 y = log2 x + log2 16
log2 y = log2 16x
y=16x-------(ii)
solving simultneously we get x=3/22, y=24/11(improper fraction forn).
massimo
i think his onemay be right and mine wrong but i routed bothj sides to get thjat answer, how did it go anyway ?? which answers were right
both of your answers were right, i jsut wasnt sure what you'd done
everyone seems to do these differently
as it turned out the whole class struggled, so she was only marking Q 1 - 6 and given us another week for 7-12
thanks guys
it wont let me give rep for some reason
it says ive given too much in the last 24hours, but ive only given one
what about this one
log3 (2-3x) = log9 (6x^2-19x+2)
log3 (2-3x) = log3 (6x^2 -19x+2) / log3 (9) changing bases
log3 (2-3x) x log3 (9) = log3 (6x^2 -19x +2) take denominator to other side
log 3 (11-3x) = log3 (6x^2 -19x + 2 ) multiplication rule
11-3x = 6x^2 -19x +2 same base so equal
6x^2 -16x-9 = 0
do you need anymore help by any chance ??
did you find out the answer to that and did you get what i did earlier about the routing to get rid of the square ??
log3 (2-3x) = log9 (6x^2-19x+2)
log3 (2-3x) = log3 (6x^2 -19x+2) / log3 (9) changing bases
log3 (2-3x) x log3 (9) = log3 (6x^2 -19x +2) take denominator to other side
log 3 (11-3x) = log3 (6x^2 -19x + 2 ) multiplication rule
11-3x = 6x^2 -19x +2 same base so equal
6x^2 -16x-9 = 0
do you need anymore help by any chance ??
did you find out the answer to that and did you get what i did earlier about the routing to get rid of the square ??
massimo
what about this one
log3 (2-3x) = log9 (6x^2-19x+2)
log3 (2-3x) = log3 (6x^2 -19x+2) / log3 (9) changing bases
log3 (2-3x) x log3 (9) = log3 (6x^2 -19x +2) take denominator to other side
log 3 (11-3x) = log3 (6x^2 -19x + 2 ) multiplication rule
11-3x = 6x^2 -19x +2 same base so equal
6x^2 -16x-9 = 0
log3 (2-3x) = log9 (6x^2-19x+2)
log3 (2-3x) = log3 (6x^2 -19x+2) / log3 (9) changing bases
log3 (2-3x) x log3 (9) = log3 (6x^2 -19x +2) take denominator to other side
log 3 (11-3x) = log3 (6x^2 -19x + 2 ) multiplication rule
11-3x = 6x^2 -19x +2 same base so equal
6x^2 -16x-9 = 0
log3 (2-3x) x log3 (9) = log3 (6x^2 -19x +2) take denominator to other side
log 3 (11-3x) = log3 (6x^2 -19x + 2 ) multiplication rule
Multiplication law
Log a xy = log a x + log a y
log 3 (2-3x) x log 3 (9) doesnt = log 3 (11-3x)
log 3 (2-3x) + log 3 (9) does = log 3 (11-3x)
Heres how you do it
log3 (2-3x) = log9 (6x^2-19x+2)
log3 (2-3x) = log3 (6x^2 -19x+2) / log3 (9) changing bases
log3 (2-3x) = 1/2[log3 (6x^2 -19x+2) ] --- Log 3 (9) = 2
2log3 (2-3x) = log3 (6x^2 -19x+2) ---- multiply both sides by 2
log3 (2-3x)^2 = log3 (6x^2 -19x+2)
log3 (2-3x)^2 - log3 (6x^2 -19x+2) = 0
log 3 (2-3x)^2/(6x^2 - 19x+2) = 0 ------ Log a (x/y) = log a x - log a y (division law)
Put this equation into y = a^x
(2-3x)^2/(6x^2 - 19x+2) = 3^0
9x^2 - 12x + 4/6x^2 -19x +2 = 1
9x^2 -12x+4 = 6x^2 -19x + 2
3x^2 + 7x + 2 = 0
(3x+1)(x+2) = 0
x = -1/3 x = -2
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