I need help solving this logarithmic equation: ln x + ln (x+3) = 1.
I know ln x + ln (x+3)=ln [x(x+3)] and that is ln x^2 + 3x when broken down. Then you have to get rid of ln by taking the natural base e of both sides so in the end you have x^2+3x = e^(1). That's when I got stuck.
You are on the right track. Remember that e^1 = e = 2.718..
You just need to use the quadratric equation to solve
x^2 + 3x -2.718 = 0
since it cannot be factored easily.
x = [-3 +/- sqrt (19.87)]/2
= 1.458/2 or -7.458/2 = 0.729 or -3.729
x+5=4
To solve the logarithmic equation ln x + ln (x + 3) = 1, you made a good start by combining the logarithms using the logarithmic identity ln(a) + ln(b) = ln(ab). So, ln x + ln (x + 3) becomes ln [x(x + 3)].
Next, you correctly simplified ln [x(x + 3)] to ln (x^2 + 3x). But instead of trying to eliminate the natural logarithm by taking the natural base e (approximately 2.71828) of both sides, you need to use the property of logarithms that states if ln(a) = b, then e^b = a.
In your equation, ln (x^2 + 3x) = 1, you can rewrite it as e^1 = x^2 + 3x. Evaluating e^1 gives you e = x^2 + 3x.
From there, you want to solve for x. Rearranging the equation, you have x^2 + 3x - e = 0. Notice that this is now a quadratic equation in terms of x.
At this point, you can solve the quadratic equation using various methods such as factoring, completing the square, or using the quadratic formula.
Once you have found the solutions for x, don't forget to check if they are valid by substituting them back into the original equation. Since the natural logarithm ln is only defined for positive numbers, any solution that makes x or (x + 3) negative is considered extraneous and should be discarded.
Hope this explanation helps you solve the logarithmic equation!