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#### melanie.dee

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« on: November 06, 2007, 09:26:46 pm »
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this one really is simple, but i dont know how to do anything involving e

anyway its asking for the x coordinate of intersection of the graphs of 2 equations. the equations are..

e^2x and 6 - e^x

there is a graphy, but i dont think you need it to solve that.

is it just when the two equations equal eachother. but then, how do i transpose that to find x, cos i dont know what you're allowed to do with e, and the rules and such

e^2x = 6 - e^x .. or is this completely retarded?

then how do i finish it

#### enwiabe

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« Reply #1 on: November 06, 2007, 09:31:20 pm »
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Yup you're pretty much there!

e^2x = 6 - e^x

e^2x + e^x - 6 = 0

Let A = e^x

so A^2 + A - 6 = 0
A_______-2
A_______+3

So A = 2, or A = -3

But A = e^x

so e^x = 2 or e^x = -3

but e^x is always positive so discard the solution e^x = -3
now go with e^x = 2

and x = log_e(2) #### costargh

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« Reply #2 on: November 06, 2007, 09:35:22 pm »
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^^^^^^^^^^^
That kinda makes sense lol. Still confuses me
Heres one for anyone to have a go at...

Solve the equation sin (x) = square root of 3 cos (x) for x (wierd symbol that i think means all elements of ) [- pie , pie], giving exact values in terms of pie

#### melanie.dee

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« Reply #3 on: November 06, 2007, 09:45:50 pm »
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Quote from: "enwiabe"
Yup you're pretty much there!

e^2x = 6 - e^x

e^2x + e^x - 6 = 0

Let A = e^x

so A^2 + A - 6 = 0
A_______-2
A_______+3

So A = 2, or A = -3

But A = e^x

so e^x = 2 or e^x = -3

but e^x is always positive so discard the solution e^x = -3
now go with e^x = 2

and x = log_e(2) ahhhh thanks a bunch, i sat there trying to work it out in strange strange ways cos i totally didnt notice that it was quadratic. gar idiot. thanks for the explanation!

#### melanie.dee

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« Reply #4 on: November 06, 2007, 09:53:01 pm »
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Quote from: "costargh"
^^^^^^^^^^^
That kinda makes sense lol. Still confuses me
Heres one for anyone to have a go at...

Solve the equation sin (x) = square root of 3 cos (x) for x (wierd symbol that i think means all elements of ) [- pie , pie], giving exact values in terms of pie

pie haha

yeh this was on the sample exam hey

sin (x) = root 3 x cos (x)

sin(x)/cos(x) = root3

tan(x) = root3

my problem here is that i dont know the exact values off by heart haha :oops:

but look it up in your text book

#### Matt89

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« Reply #5 on: November 06, 2007, 09:53:10 pm »
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Quote from: "costargh"
^^^^^^^^^^^
That kinda makes sense lol. Still confuses me
Heres one for anyone to have a go at...

Solve the equation sin (x) = square root of 3 cos (x) for x (wierd symbol that i think means all elements of ) [- pie , pie], giving exact values in terms of pie

Divide both sides by cos(x) and you get....
tan(x)=square root 3
x = -2pi/3 and pi/3

EDIT:

Ah, too late

But exact values are very important for the exam, so you should learn them

#### costargh

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« Reply #6 on: November 06, 2007, 09:56:09 pm »
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Ohh yehhh damn I should know that :@