C1 Edexcel Jan2010 Question! Please help!

[~ A.F ~]

whats the "?" in ur question?

[halosh92]

Giving your answers in terms of pie, solve the equation
3 tan2 thita - 1=0
for thita in the interval ?pie ? thita ? pie

[halosh92]

(a) Given that t = log3 x, find expressions in terms of t for
(i) log3 x2,
(ii) log9 x.
(b) Hence, or otherwise, find to 3 significant figures the value of x such that
log3
x2 ? log9 x = 4.

how to do part aii

[nid404]

Giving your answers in terms of pie, solve the equation
3 tan2 thita - 1=0
for thita in the interval -pie < thita < pie

tan 2 theta= 1/3
2 theta= tan inverse of 1/3
2 theta= 18.4
theta=9.22
period of tan 2theta= +-90 [2theta right...so it will be half the tan theta period...1/2 of 180]
       = 9.22, 99.22, -80.78, -170.78

[T.Q]

(a) Given that t = log3 x, find expressions in terms of t for
(i) log3 x2,
(ii) log9 x.
(b) Hence, or otherwise, find to 3 significant figures the value of x such that
log3
x2 ? log9 x = 4.

how to do part aii

for part aii

change the base

and make it 3

so it will be like this   (log3 x) /(log3 9)

so log3 x = t

and log3 9 = 2

so the answer is (1/2 t)

[astarmathsandphysics]

(a) Given that t = log3 x, find expressions in terms of t for
(i) log3 x2,
(ii) log9 x.
(b) Hence, or otherwise, find to 3 significant figures the value of x such that
log3
x2 ? log9 x = 4.

ai)2t
ii)log9 x =lig3 x^(1/2) =1/2 t
retype qb

[halosh92]

for part aii

change the base

and make it 3

so it will be like this   (log3 x) /(log3 9)

so log3 x = t

and log3 9 = 2

why cant we use the inverse rule here....like make it 1/log x3 ...............where log x3=1/t
which rule did u use there exactly....... how did u derive the base as 3 in (log3 x) /(log3 9)

so the answer is (1/2 t)

[T.Q]

i used this rule  

so log9 x = log x /log 9  , then u put any  base u want , i chose 3

[halosh92]

i used this rule  

so log9 x = log x /log 9  , then u put any  base u want , i chose 3

thankyou

[halosh92]

people
could anyone do q7c
plzzz urgentttt

[astarmathsandphysics]

a=129 r=114/120=0.95
ar^(n-1)<60
120*0.95^n<60
log120 +n log 0.95 < log 60
n?(log 60 -log 120)/log 0.95 =13.52
n=14

[halosh92]

Q4
plz explain in detailsss
thx alotttt

[astarmathsandphysics]

or you could just keep miultiplying 120 by 0.95 until your answer is less that 60

[halosh92]

or you could just keep miultiplying 120 by 0.95 until your answer is less that 60

why do we do that?
could u be more specific plz?

[nid404]

Q4
plz explain in detailsss
thx alotttt

Do you need the a) part as well? (i mean the curve)

b) sin^-10.35= x-30
    x=20.5^o+30
     =50.5^o