Solving exponential equations (2 Viewers)

[Hypem]

So I came across a question in the 2007 HSC that required you to solve an exponential equation.

I haven't learnt this at all...more the reason to love Maths in Focus. Seriously thinking about picking up a better textbook, it's trash.

Anyway, so if anyone could teach me how to solve this it'd be great!

http://puu.sh/2CWvg.png

Thanks!

 

[Nws m8]

Pull e^x out mate :


=> 2e^x.e^x - e^x = 0
=> e^x (2e^x -1)= 0

and you solve the rest

 

[Nws m8]

And as far as textbook is concerned, MATH in FOcus is good for understanding and brushing up your basics, for HSC preparation, do as many past papers as you can !

 

[Hypem]

Yeah so that gets me e^x = 1/2, but i don't know what to do to get x

 

[Aysce]

Yeah so that gets me e^x = 1/2, but i don't know what to do to get x

You take the natural log of e^x and 1/2 so:

e^x = 1/2

ln(e^x) = ln(1/2)

x = ln(1/2) = -ln(2)

 

[Nws m8]

Yeah so that gets me e^x = 1/2, but i don't know what to do to get x

Take log on both sides

Edit: Aysce has already done that lol

 

[Hypem]

You take the natural log of e^x and 1/2 so:

e^x = 1/2

ln(e^x) = ln(1/2)

x = ln(1/2) = -ln(2)

Ohhhhh so basically:

ln(e^x) is xln(e), which is x *1 = x

So x = ln(1/2)
= ln1 - ln2
= 0 - ln(2)
= -ln(2)

Does anyone have heaps of these so I can practice?

 

[Aysce]

Ohhhhh so basically:

ln(e^x) is xln(e), which is x *1 = x

So x = ln(1/2)
= ln1 - ln2
= 0 - ln(2)
= -ln(2)

Does anyone have heaps of these so I can practice?

You should find more basic questions like these in Maths in Focus and then progress to harder levels such as past papers and the Cambridge textbook. Also instead of breaking up ln(1/2) into ln(1) - ln(2), it's a little bit quicker to see that ln(1/2) = ln(2^-1) = -ln(2).

 

[Hypem]

lol Maths in Focus doesn't have any of these though... that's why I couldn't do it since we never learnt how to.

Thanks for helping though

 

[Aysce]

lol Maths in Focus doesn't have any of these though... that's why I couldn't do it since we never learnt how to.

Thanks for helping though

No, I am very sure Maths in Focus at least has these types of questions.

 

[Hypem]

Oh you're right, it's in Ex 4.6, I think I skipped over questions 3+ and forgot to do them.
Guess I'll do them now lol.

 

[Nws m8]

Ohhhhh so basically:

ln(e^x) is xln(e), which is x *1 = x

So x = ln(1/2)
= ln1 - ln2
= 0 - ln(2)
= -ln(2)

Does anyone have heaps of these so I can practice?

Do this (I just made it up) :

e^4x + e^2x - 2 = 0 (very simple question)

 

[Hypem]

x = 0 from just looking at it

but doing that the long way -- how do you factorise properly?

 

[Nws m8]

x = 0 from just looking at it

but doing that the long way -- how do you factorise properly?

BIG Hint: it's a reducible quadratic

 

[Hypem]

Where do you put the 2 though?

 

[Hypem]

Oh wait, you mean just let p = e^2x or something?

so p^2 + p -2 = 0

P: -2
S: 1
F: 2,-1

p^2 + 2p -p - 2 = 0
p(p+2) -(p+2) = 0
(p-1)(p+2) = 0
p = 1, -2

e^x = 1, e^x = -2

x = ln(1), x= ln(-2)
x = 0, x = imaginary?

 

[Drongoski]

Does anyone have heaps of these so I can practice?

Just do all the questions in (based on the old 2nd edition - 1997) Signpost Maths Yr 10 (Advanced):

14:04 - Simple Exponential Equations

14:05 - Logarithms

14:06 - Logarithms & Exponential Graphs

14:07 - Laws of Logarithms

14:08 - Further Exponential Equations

If you can do them all then your knowledge of Logarithms would be good enough for MX1.

 

[Nws m8]

Oh wait, you mean just let p = e^2x or something?

so p^2 + p -2 = 0

P: -2
S: 1
F: 2,-1

p^2 + 2p -p - 2 = 0
p(p+2) -(p+2) = 0
(p-1)(p+2) = 0
p = 1, -2

e^x = 1, e^x = -2

x = ln(1), x= ln(-2)
x = 0, x = imaginary?

Mistake

 

[Hypem]

Oh fuck woops

You still get x = 0 and imaginary though

 

[Hypem]

Btw when you pull the power out, is it 2x = .... or x^2 = ....?