vt33 said:Can anyone help me with this natural log?
ln x + ln(x+1) = 7
I've tried putting it in quadratic form but all I get is x^2 + x - 7=0
which doesn't seem to factor. What am I doing wrong?
vt33 said:well, thanks but, I've never actually seen this type of equation. What do I do with the -e^7, or what I mean is how do I factor that? Actually, should I just use the quadratic formula?
The first step is to combine the two logarithms using the product rule, which states that ln(a) + ln(b) = ln(ab). This gives us ln(x(x+1)) = 7.
We can isolate x by taking the inverse function of ln, which is e^x. This gives us e^ln(x(x+1)) = e^7, which simplifies to x(x+1) = e^7.
Yes, we can use the distributive property to expand x(x+1) to x^2 + x, giving us the quadratic equation x^2 + x - e^7 = 0.
We can use the quadratic formula x = (-b ± √(b^2-4ac)) / 2a, where a = 1, b = 1, and c = -e^7. This gives us two possible solutions for x: x = (-1 ± √(1+4e^7)) / 2.
Yes, since we are taking the natural logarithm of x and x+1, both values must be positive. Additionally, the solutions obtained from the quadratic formula may be complex numbers, so we must also check if they satisfy the original equation or not.