.5ln(x+3) - lnx=0
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~ :)
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To solve the equation .5ln(x+3) - ln(x) = 0, you can follow these steps:
Step 1: Combine the logarithms using the rule ln(a) - ln(b) = ln(a/b).
.5ln(x+3) - ln(x) = ln((x+3)^.5 / x) = 0
Step 2: Apply the exponential function e^x to both sides to eliminate the logarithm.
e^(ln((x+3)^.5 / x)) = e^0
(x+3)^.5 / x = 1
Step 3: Square both sides of the equation.
((x+3)^.5 / x)^2 = 1^2
(x+3) / x^2 = 1
Step 4: Cross multiply to get rid of the fraction.
x+3 = x^2
Step 5: Rearrange the equation to standard quadratic form.
x^2 - x - 3 = 0
Step 6: Solve the quadratic equation using factoring, completing the square, or using the quadratic formula.
Since the equation does not factor easily, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -1, and c = -3:
x = (-(-1) ± √((-1)^2 - 4(1)(-3))) / (2(1))
Simplifying further:
x = (1 ± √(1 + 12)) / 2
x = (1 ± √(13)) / 2
Therefore, the solutions to the equation .5ln(x+3) - ln(x) = 0 are:
x = (1 + √(13)) / 2
x = (1 - √(13)) / 2