Solve for x...?

Question

I'm completely lost on these 2...

1. ln(5-2x)= -3

2. 2lnx=ln2 + ln(3x-4)

Answer 1

Question 1
(5 - 2x) = e^(-3)
5 - 2x = 0.05
2x = 4.95
x = 2.475

Question 2
ln ( x ² ) = ln [ 2 (3x - 4) ]
x ² = 2 (3x - 4)
x ² = 6x - 8
x ² - 6x + 8 = 0
(x - 4) (x - 2) = 0
x = 4 , x = 2

Answer 2

e^ ln is the same as e^log base e which is 1. Use a scientific calculator to solve.

1. e^ln(5-2x) = e^ (-3)
5-2x = e^-3
2x= 5-e^-3
x= 2.48

2. e^ 2lnx = e^ ln(3x-4)
(2lnx is the same as ln x^2)
x^2 = 3x-4
x^2-3x+4= 0
(x-4)(x+1)=0
x=4
(x=-1) is not a possible solution because base e to some power can't give you an answer of -1.

Answer 3

ln is a natural logarithm
So you take e^ln(1) >>>1
So here
1) we get 5-2x= e^-3 which becomes, x= [(e^-3) -5]/2


2) e^2 (x) =2 + 3x-4 or 3x -2
which can become (e^2 -3)x= -2
so x= -2/ (e^2 -3)

Answer 4

#1
5-2x = e^(-3)
-2x = e^(-3)-5
x = -1/2 * (e^(-3)-5)
x = 2.475106