solve. provide the exact answer when possible. when not possible round to 4 decimal place.
a) ln(x) +2 = ln(x+3)
(Nina, please be sure to use Ask a New Question instead of piggybacking on another student's post.)
ln(x) +2 = ln(x+3)
replace 2 with ln e^2
ln(x) +ln e^2 = ln(x+3)
ln (xe^2) = ln(x+3)
anti-ln both sides
xe^2 = x+3
xe^2 - x = 3
x(e^2 - 1) = 3
x = 3/(e^2 - 1)
ln x - ln (x+3) = ln [ x/(x+3) ]
so ln [ x/(x+3) ] = -2
remember e^lna = a
so
x/(x+3) = e^-2
now you do it
To solve the equation ln(x) + 2 = ln(x + 3), we can use the properties of logarithms.
Step 1: Combine the logarithms
Since both logarithms have the same base (natural logarithm, ln), we can use the property that ln(a) + ln(b) = ln(a * b). Applying this property, we can rewrite the equation as ln(x(x + 3)) + 2 = 0.
Step 2: Remove the natural logarithm
To remove the natural logarithm, we need to exponentiate both sides of the equation with the base e (Euler's number). This will cancel out the natural logarithm and leave us with x(x + 3) = e^(-2).
Step 3: Solve the quadratic equation
Expanding x(x + 3), we get x^2 + 3x. Our equation becomes x^2 + 3x = e^(-2).
To solve this quadratic equation, we can set it equal to zero and then apply the quadratic formula:
x^2 + 3x - e^(-2) = 0
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = 3, and c = -e^(-2). Plugging in these values, our quadratic formula becomes:
x = (-3 ± √(3^2 - 4(1)(-e^(-2)))) / (2(1))
Simplifying further, we have:
x = (-3 ± √(9 + 4e^(-2))) / 2
This is the exact solution to the equation. If the exact solution is not needed, you can substitute the value of e (approximately 2.7183) and calculate the approximate values for further steps.
Note: If you need a specific value for x, you can use a calculator or round the square root to 4 decimal places and evaluate the expression.