Use logarithms to solve the equation (http://latex.codecogs.com/gif.latex?\bg_black%207^x%20=%202^x^-^1), giving the value of x correct to 3 S.F. (5 marks)
pls kthx
24
Quote from: Toast on April 20, 2013, 12:03:33 PM
24
giving the value of x correct to 3 S.F. (5 marks)
ok???!??!
sry its 8
log both sides:
log (7^x) = log (2^x-1)
take the x and x-1 outside log:
x log7 = (x-1) log 2
multiply out:
x log 7= x log 2 - log 2
x log 7 - x log 2 = - log 2
x (log 7 - log 2) = - log 2
x = - log 2 / (log 7 - log 2) = -0.553
i winz
i was joking, i don't even know what logarithms is.
Quote from: Toast on April 20, 2013, 12:35:58 PM
i was joking, i don't even know what logarithms is.
they dont teach that in kindergarten :D
Quote from: Iridion on April 20, 2013, 12:41:14 PM
Quote from: Toast on April 20, 2013, 12:35:58 PM
i was joking, i don't even know what logarithms is.
they dont teach that in kindergarten :D
ya what he said
42.6
Quote from: lasercut on April 20, 2013, 12:32:04 PM
log both sides:
log (7^x) = log (2^x-1)
take the x and x-1 outside log:
x log7 = (x-1) log 2
multiply out:
x log 7= x log 2 - log 2
x log 7 - x log 2 = - log 2
x (log 7 - log 2) = - log 2
x = - log 2 / (log 7 - log 2) = -0.553
i winz
seems most legit. here is my thanks priz: WINNER (http://eshrooms.teamdead.net/)
lasercut is the only person who solved that question correctly...
logs are pretty ez bro.
there be three r00lz:
1) if you are adding to logs, you can combine the two logs by multiplying by the numbers "inside"
ex: log 2 + log 3 = log (2*3) = log 6
2) if you are subtracting logs, you can combine two logs by dividing by the numbers "inside"
ex: log 6 - log 3 = log (6/3) = log 2
3) if you have an exponent in a log, you can move that exponent to the front and distribute it
ex: log 6^(x-1) = (x-1) * log 6 = (x * log 6) - log 6
source: i did logs in algebra II and got a 98% in the class
^ got th low down br0 was jus harrrd in the brain
Quote from: antisickness on April 20, 2013, 01:32:21 PM
Quote from: lasercut on April 20, 2013, 12:32:04 PM
log both sides:
log (7^x) = log (2^x-1)
take the x and x-1 outside log:
x log7 = (x-1) log 2
multiply out:
x log 7= x log 2 - log 2
x log 7 - x log 2 = - log 2
x (log 7 - log 2) = - log 2
x = - log 2 / (log 7 - log 2) = -0.553
i winz
seems most legit. here is my thanks priz: WINNER (http://eshrooms.teamdead.net/)
EPILEEEPSYYYYYY
i do maths.
Quote from: kulvinder on April 20, 2013, 05:23:12 PM
If x=2 when y =3 then y=? when x=5? ::)
ah, gr. 7. wat a wonderful year
z fukn high thinkin he can maths. fuk of ur brein is rustey radiator
y dont ppl do my homework