Solve for x:
ln(x+1)+ 2 = x^2 – 8x + 12
wow, that's a nasty one
do you have a graphing calculator?
if so, graph y = ln(x+1) and
y = x^2 - 8x + 10
zoom in on their intersection point.
I don't have one of those, but with a few trial and error calculations on my oldfashioned machine
I got x = appr 1.38
can't use a graphing calculator (not allowed). I'm trying to find the area between curves, if that helps.
I ran your equation through a Newton's Method program I have, and I got
x = 1.3795 and
x = 6.83885
If you are going to find area, you might have to integrate ln(x+1)....
hint: take the derivative of
g(x) = (x+1)ln(x+1) and see what you get.
mckendrick?
To solve for x in the equation ln(x+1) + 2 = x^2 – 8x + 12, we need to eliminate the natural logarithm and rearrange the equation.
Step 1: Subtract 2 from both sides of the equation:
ln(x+1) = x^2 - 8x + 10
Step 2: Rewrite the equation without the natural logarithm:
e^(ln(x+1)) = e^(x^2 - 8x + 10)
Step 3: Simplify the left side of the equation using the properties of logarithms:
x + 1 = e^(x^2 - 8x + 10)
Step 4: Convert the right side of the equation to one exponent using the property e^(a+b) = e^a * e^b:
x + 1 = e^x^2 * e^(-8x) * e^10
Step 5: To solve for x, we can use numerical or graphical methods, as the equation does not have a simple algebraic solution. One way to approach this is to plot the graph of y = x + 1 and y = e^x^2 * e^(-8x) * e^10 and observe their intersection points. Another method is to use iterative numerical methods such as the Newton-Raphson method or the bisection method to approximate the value of x that satisfies the equation. These numerical methods involve repeatedly calculating values based on the given equation to get closer to the solution.
Note: It's important to remember that not all equations can be solved with simple algebraic techniques, and numerical methods are often required for finding approximate solutions.