lnx+ ln (x+3)= 2ln2
Using your laws of logs
ln[x(x+3)] = ln 2^2
x(x+3) = 4
x^2 + 3x - 4 = 0
it factors and solves easily , remember that for ln x to be defined, x > 0
To solve the equation lnx + ln(x+3) = 2ln2, we can use logarithmic properties and algebraic manipulation. Here's a step-by-step solution:
Step 1: Apply the rules of logarithms to simplify the equation.
Using the rule of logarithms, ln(a) + ln(b) = ln(a*b), we can rewrite the equation as:
ln(x(x+3)) = ln(2^2)
Step 2: Simplify the right side of the equation.
Since ln(2^2) = ln(4), we rewrite the equation as:
ln(x(x+3)) = ln(4)
Step 3: Eliminate the natural logarithm and convert it into exponential form.
If ln(a) = ln(b), then a = b. Using this property, we can write:
x(x+3) = 4
Step 4: Expand and rearrange the equation.
Multiply x by x+3:
x^2 + 3x = 4
Step 5: Move all terms to one side of the equation.
x^2 + 3x - 4 = 0
Step 6: Factorize the quadratic equation.
After factoring, we get:
(x + 4)(x - 1) = 0
Step 7: Solve for x.
Setting each factor to zero, we have two separate equations:
x + 4 = 0 or x - 1 = 0
Solving for x, we find:
x = -4 or x = 1
So, the solutions to the equation lnx + ln(x+3) = 2ln2 are x = -4 and x = 1.