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Benarism

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Nov 4, 2013
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I have this problem. The question is to find the exact solution of e^x + 3e^-x = 4.

Automatically I thought to rewrite this as e^x + 3/e^x = 4. Then if i multiply through by e^x, I will have a quadratic in terms of e^x. So this will give me e^2x - 4e^x + 3 = 0. Factorising this will give me
(e^x - 1) (e^x - 3) = 0. Therefore e^x = 1 or e^x = 3. Hence x = ln1 = 0 or x = ln3. So i have two solutions.

However, I had a friend of mine who went for this approach.
So they took the natural log of all terms, so
lne^x + 3lne^-x = ln4
x -3x = ln4
-2x = ln4
x = -ln4/2

as you can see you only get one solution with my friends approach. And it is not the same solution i got. Can someone please tell me why my friends approach is not right? Could you please explain it fully. I am really confused right now. Thank you
 
Benarism said:
I have this problem. The question is to find the exact solution of e^x + 3e^-x = 4.

Automatically I thought to rewrite this as e^x + 3/e^x = 4. Then if i multiply through by e^x, I will have a quadratic in terms of e^x. So this will give me e^2x - 4e^x + 3 = 0. Factorising this will give me
(e^x - 1) (e^x - 3) = 0. Therefore e^x = 1 or e^x = 3. Hence x = ln1 = 0 or x = ln3. So i have two solutions.

However, I had a friend of mine who went for this approach.
So they took the natural log of all terms, so
lne^x + 3lne^-x = ln4 X
x -3x = ln4
-2x = ln4
x = -ln4/2

as you can see you only get one solution with my friends approach. And it is not the same solution i got. Can someone please tell me why my friends approach is not right? Could you please explain it fully. I am really confused right now. Thank you
Click to expand...
You are right. Tell your friend he can NOT take the logarithm of a sum. The sum of logarithms is the logarithm of the product.
 
Benarism said:
I have this problem. The question is to find the exact solution of e^x + 3e^-x = 4.

Automatically I thought to rewrite this as e^x + 3/e^x = 4. Then if i multiply through by e^x, I will have a quadratic in terms of e^x. So this will give me e^2x - 4e^x + 3 = 0. Factorising this will give me
(e^x - 1) (e^x - 3) = 0. Therefore e^x = 1 or e^x = 3. Hence x = ln1 = 0 or x = ln3. So i have two solutions.

However, I had a friend of mine who went for this approach.
So they took the natural log of all terms, so
lne^x + 3lne^-x = ln4
x -3x = ln4
-2x = ln4
x = -ln4/2

as you can see you only get one solution with my friends approach. And it is not the same solution i got. Can someone please tell me why my friends approach is not right? Could you please explain it fully. I am really confused right now. Thank you
Click to expand...

You can back-substitute your solution into the given equation and check!

for x = 0

e^x + 3/e^x = e^0 + 3/e^0 = 1 + 3/1 = 4 ...............original equation checks

for x = ln(3)

e^x + 3/e^x = e^[ln(3)] + 3/e^[ln(3)] = 3 + 3/3 = 4 ...............original equation checks

So your solutions are correct.

You can similarly show that the other "alleged" solution does not "check".
 
Benarism said:
I have this problem. The question is to find the exact solution of e^x + 3e^-x = 4.

Automatically I thought to rewrite this as e^x + 3/e^x = 4. Then if i multiply through by e^x,
I will have a quadratic in terms of e^x. So this will give me e^2x - 4e^x + 3 = 0.
Click to expand...

Put grouping symbols around the first exponent:

e^(2x) - 4e^x + 3 = 0.
 
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