Find the taylor series for f(x) = ln x centered at a = 2

1. Find the taylor series for f(x) = ln x centered at a = 2.[Assumethat f has a power series expansion. Do not show thatRn(x)–>0.]
2. Solve each of the following equations: (a) (b) 5 – 2x = 17 6x +1 = 4(3x – 7) 2(x – 4) + = 7 5 10 2. – 71 = 12 (c) (d)

The Correct Answer and Explanation is :

Let’s break down the tasks one by one:

1. Find the Taylor Series for ( f(x) = \ln(x) ) centered at ( a = 2 ):

To find the Taylor series of a function ( f(x) ) around a point ( a ), we use the general formula for a Taylor series:

[
f(x) = f(a) + f'(a)(x – a) + \frac{f”(a)}{2!}(x – a)^2 + \frac{f^{(3)}(a)}{3!}(x – a)^3 + \dots
]

For ( f(x) = \ln(x) ), the derivatives of ( f(x) ) are:

1. ( f(x) = \ln(x) )
2. ( f'(x) = \frac{1}{x} )
3. ( f”(x) = -\frac{1}{x^2} )
4. ( f^{(3)}(x) = \frac{2}{x^3} )
5. ( f^{(4)}(x) = -\frac{6}{x^4} )
   (and so on)

Now, we evaluate these derivatives at ( x = 2 ):

1. ( f(2) = \ln(2) )
2. ( f'(2) = \frac{1}{2} )
3. ( f”(2) = -\frac{1}{4} )
4. ( f^{(3)}(2) = \frac{2}{8} = \frac{1}{4} )
5. ( f^{(4)}(2) = -\frac{6}{16} = -\frac{3}{8} )

Therefore, the Taylor series expansion of ( \ln(x) ) centered at ( a = 2 ) is:

[
\ln(x) = \ln(2) + \frac{1}{2}(x – 2) – \frac{1}{8}(x – 2)^2 + \frac{1}{24}(x – 2)^3 – \frac{3}{192}(x – 2)^4 + \dots
]

This series can be truncated depending on the desired level of accuracy.

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2. Solve the following equations:

(a) ( 5 – 2x = 17 )

We solve for ( x ):

[
5 – 2x = 17
]
[
-2x = 17 – 5
]
[
-2x = 12
]
[
x = -6
]

Thus, the solution to the first equation is ( x = -6 ).

(b) ( 6x + 1 = 4(3x – 7) )

Distribute the ( 4 ) on the right side:

[
6x + 1 = 12x – 28
]

Now, move all terms involving ( x ) to one side and constants to the other:

[
6x – 12x = -28 – 1
]
[
-6x = -29
]
[
x = \frac{29}{6}
]

Thus, the solution to the second equation is ( x = \frac{29}{6} ).

(c) ( 2(x – 4) + 5 = 7 )

Distribute the ( 2 ):

[
2x – 8 + 5 = 7
]
[
2x – 3 = 7
]
[
2x = 7 + 3
]
[
2x = 10
]
[
x = 5
]

Thus, the solution to the third equation is ( x = 5 ).

(d) ( -71 = 12 )

This equation seems incorrect or incomplete as it stands. If you meant to solve for ( x ), or if there was an error in the equation provided, please clarify.

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