Calculus...solve for x.?

Question

1-e^(2x)=3 and ln(2x+1)^2=12


details for both problems please

Answer 1

Problem 1:
Collect constants on the right and mult by -1
e^(2x) = 2
Take nat logs of each side. 2x = ln 2
Since ln 2 = 0.693, then x = 0.693/2

Same deal, but in reverse.
Take both sides to eth power:
(2x+1)^2 = e^12
Evaluate e^12, take square root, call result A.
then 2x+1 = A or x = (A-1)/2

Answer 2

Some formulas
ln(e^x) = x for all real x
ln (a)^b = b ln (a)
e^(ln(x)) = x if x > 0


1-e^(2x)=3
1 - 3 = e^(2x)
-2 = e^(2x)
multiply both sides by ln
ln (-2) = ln (e^(2x))
ln (-2) = 2x (because ln and e cancel each other)
There is no solution for ln (-2). Ln only have solution for positive values.
So there is no solution for x



ln(2x + 1)^2=12
take the exponent on both sides
e^{ln (2x + 1)^2} = e^(12) (because ln and e cancel each other)
(2x + 1)^2 = e^(12)
take sqrt on both sides
(2x + 1)^(2 / 2) = e^(12 / 2)

2x + 1 = e^(6)
x = [e^(6) - 1] / 2

Answer 3

1 - e^(2x) = 3
1 - 3 = e^(2x)
-2 = e^(2x)
ln (-2) = ln [e^(2x)]
ln (-2) = 2x
Can't take ln of negative. Therefore, No Solution.

ln (2x + 1)^2 = 12
(2x + 1)^2 = e^12
4x^2 + 4x + 1 - e^12 = 0
I would graph it on my TI and perform Calc Zero to find the two potential solutions. Remember to check the answers. One or both may be extraneous.