.5ln(x+3) - lnx=0

Answer 1

first we can transpose the - lnx to the right side of equation,
0.5 ln (x+3) = ln x
recalling some laws of exponents, we can rewrite 0.5 ln (x+3) as
ln (x+3)^0.5
therefore
ln (x+3)^0.5 = ln x
now we can equate the terms inside the ln:
(x+3)^(1/2) = x
square both sides:
x+3 = x^2
x^2 - x - 3 = 0
to solve this, we use quadratic formula:
x = [-b +- sqrt(b^2 - 4*a*c)]/(2*a)
where
a = numerical coefficient of x^2
b = numerical coefficient of x
c = the constant
substituting:
x = [-(-1) +- sqrt((-1)^2 - 4(1)(-3))]/(2(1))
x = [1 +- sqrt(1 + 13)]/2
simplifying further,
x = -1.303
x = 2.303

note tha if we substitute back both of these values, we can see that x = -1.303 is extraneous because ln (-1.303) is undefined, therefor we only get the positive value:

x = 2.303

hope this helps~ :)

Answer 2

Thank you sooo much! I really appreciate it!!