Question

25 POINTS

Answer 1

Do you want a equation?

Answer 2

Answer:

$x_{5}=0.85$

Step-by-step explanation:

We can use Newton's method to solve this kind of problems.

It is a method of computing fixed points of iterated functions, the equation is given by:

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$

So, first of all we need to identify the function.

We know that our equation is: $-2x+6=-(2/3)^{x}+5$

Let's simplify and rewrite this equation using the natural log.

$ln(2x-1)=xln(2/3)$ and if we put all these terms on the left side we will get the function.

$f(x)=ln(2x-1)-x*ln(2/3)$

f'(x) is the first derivative of the function, Let's take the derivative.

$f'(x)=\frac{2}{2x-1}-ln(2/3)$

Now, using the graph we choose x₁ = 0.6 as a starting point. Let's recall we need to find the intersection point, so we must choose a value of x as close as possible to the intersect point.

$x_{1}=0.6$

$x_{2}=x_{1}-\frac{f(x_{1})}{f'(x_{1})}$

$x_{2}=x_{1}-\frac{ln(2*x_{1}-1)-x_{1}*ln(2/3)}{2/(2*x_{1}-1)-ln(2/3)}$

$x_{2}=0.6-\frac{ln(2*0.6-1)-0.6*ln(2/3)}{2/(2*0.6-1)-ln(2/3)}$

$x_{2}=0.73$

We need to put x₂ in the equation again and find x₃

$x_{3}=0.73-\frac{ln(2*0.73-1)-0.73*ln(2/3)}{2/(2*0.73-1)-ln(2/3)}$

$x_{3}=0.83$

$x_{4}=0.83-\frac{ln(2*0.83-1)-0.83*ln(2/3)}{2/(2*0.83-1)-ln(2/3)}$  

$x_{4}=0.85$

$x_{5}=0.85-\frac{ln(2*0.85-1)-0.85*ln(2/3)}{2/(2*0.85)-ln(2/3)}$  

$x_{5}=0.85$

Therefore the approximate solution is 0.85.

I hope it helps you!