how do i solve this
2 ln(x-3)= ln(x+5) + ln4
Use the laws of logarithm:
ln a + ln b = ln ab
2ln a = ln a²
2 ln(x-3)= ln(x+5) + ln4
=>
ln(x-3)² = ln 4(x+5)
Raise to power of e:
(x-3)² = 4(x+5)
Simplify and factor to get:
(x-11)*(x+1)=0
Can you take it from here?
yes thank you much! :)
To solve the equation 2 ln(x-3) = ln(x+5) + ln4, we can use the properties of logarithms. Here's the step-by-step solution:
Step 1: Combine the logarithms on the right side using the property ln(a) + ln(b) = ln(ab). This gives us:
2 ln(x-3) = ln(4(x+5))
Step 2: Apply the property ln(a^b) = b ln(a) to eliminate the coefficient of the logarithm on the left side. This gives us:
ln((x-3)^2) = ln(4(x+5))
Step 3: Remove the logarithm on both sides by taking the exponential of both sides with base e. This will give us the equation:
(x-3)^2 = 4(x+5)
Step 4: Expand the equation by squaring the binomial on the left side:
x^2 - 6x + 9 = 4x + 20
Step 5: Simplify the equation by combining like terms:
x^2 - 10x - 11 = 0
Step 6: Solve the quadratic equation using factoring or the quadratic formula. In this case, the equation can be factored as:
(x - 11)(x + 1) = 0
Setting each factor equal to zero gives two possible solutions:
x - 11 = 0 -> x = 11
x + 1 = 0 -> x = -1
So the solutions to the equation 2 ln(x-3) = ln(x+5) + ln4 are x = 11 and x = -1.