logarithmic functions?

Question

Solve the equation....

ln (5x + 2) + ln x = ln e

and

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

Answer 1

1.
ln(5x + 2) + ln x = ln e
ln(5x + 2) + ln x = 1
e ^ (ln(5x + 2) + ln x) = e ^ (1)
(e ^ ln(5x + 2))(e ^ ln x) = e
(5x + 2)(x) = e
5x^2 + 2x - e = 0
x = (-2 +/- sqrt(4 - (4)(5)(-e))) / 2(5)
x = (-2 +/- sqrt(4 + 20e)) / 10
x = (-2 +/- (2)sqrt(1 + 5e)) / 10
x = (-1 +/- sqrt(1 + 5e)) / 5
test answers to ensure ln of positive number
x = (-1 + sqrt(1 + 5e)) / 5

2.
ln(x + 1) / ln(x - 1) = 2
ln(x + 1) = (2)ln(x - 1)
ln(x + 1) = ln((x - 1)^2)
x + 1 = (x - 1) ^ 2
x + 1 = x^2 - 2x + 1
0 = x^2 - 3x
x(x - 3) = 0
x = 0 or x = 3
test answers to ensure ln of positive number
x = 3

Answer 2

A. ln(5x+2) + ln(x) = ln(e)
(5x+2) * x = e difference/addition of logs
5x^2+2x-e = 0
solve as a quadratic
x = 0.563794, you will get another negative result, but that won't work as you cannot take the natural log of it from your original equation.
B. ln(x+1) / ln(x-1) = 2
(x+1) / (x-1) = 2
x+1 = 2 * (x-1)
x+1 = 2x-2
3=x

Answer 3

for the first
(5x+2)x=1 but x>0
x=1+-6^(1/2)/5
but ignore -ve value
x=1+6^(1/2)/5
for the second
x+1=(x-1)^2 but x>1
x=0,3
but ignore x=0
therefore x=3