Question

What is the integer solution of the following equation
e^(2x)-3e^(x)=-2

Answer 1

solve like a quadratic

u^2 -3u +2 = 0

where u = e^x

Answer 2

Yes. Use the substitution suggested above.

But if you wish to you can directly factor it by grouping:

e^(2x) - 3e^x = -2
e^(2x) - 3e^x + 2 = 0
e^(2x) - e^x - 2e^x + 2 = 0
e^x(e^x - 1) - 2(e^x - 1) = 0
(e^x - 1)(e^x - 2) = 0
e^x = 1 or e^x = 2
solve for x.

Answer 3

Is there a reason that when you get to the 4th step there is no 3?

Answer 4

On line 3, the -3e^x is split into -e^x - 2e^x to help with the factoring. Was that your question?

Answer 5

yes that was my question. thanks:) when solving for x, I put log on both sides?

Answer 6

yes. you take natural log or ln on both sides.
Remember log or ln of 1 is 0
and ln(e) = 1

Answer 7

really lost now…

Answer 8

e^x = 1 or e^x = 2
start with e^x = 1
take ln on both sides:
ln(e^x) = ln(1)
x * ln(e) = ln(1)
x * 1 = 0
x = 0

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

x = 0 or ln(2)

Answer 9

\[\Large \ln(a)^m = m * \ln(a)\]

Answer 10

\[\Large \ln(e)^x = x * \ln(e) = x\]

Answer 11

the first one is the equation and the second one has everything plugged in?ln(a)m=m∗ln(a)
ln(e)x=x∗ln(e)=x

Answer 12

The first one is an IDENTITY. That means it is good for a whole range of a and m. Whenever we come across ln(something)^exponent the identity says it can be written as:

ln(something)^exponent = exponent * ln(something)

In the second case we make use of the identity to rewrite ln(e^x) as x * ln(e) and ln(e) = 1.

Answer 13

Examples:\[\Large \ln(5)^8 = 8 * \ln(5)\]\[\Large \ln(e)^5 = 5 * \ln(e) = 5\]

Answer 14

Thanks you very much :) am beginning to understand it more. which is a positive considering i have a midterm tomorrow.

Answer 15

cool. There are just a few logarithmic identities. And once you familiarize yourself you can breeze through problems involving logarithms.

Answer 16

you are welcome.