#### To determine

**(**a)

**To find:**

That there is exactly one root of the given equation and it lies between (2,e).

**Solution:**

Yes, we could prove that there is exactly one root of the given equation and it lies between (2,e).

**Given:** lnx=3−x

**Formulae used:** The intermediate value theorem

**Explanation:**

Consider f(x)=lnx+x−3

f(x)=lnx+x−3f′(x)=1x+1

For x>0, f′(x)=1x+1>0

The value of function at x=2, x=e are f(2)≈−0.307 and f(e)≈0.718 respectively

Therefore, f(x) is continuous on [2,e] and differentiable on (2,e)

Hence, use the formulae of intermediate value theorem ∃c ∈(2,e), such that f(c)=0, it means that there is at least one root on this interval.

Thus, as seen from the above value f(2)<0 and f(e)>0

But f′(x)>0 for x∈(2,e), it means that the function is increasing on this interval and there can be only one root.

**Conclusion:** Hence there is exactly one root lie between (2,e)

**(** **b )**

**To find:** The root of the equation to four decimal places.

**Solution:** 2.2079

**Given:** The expression is lnx=3−x

**Formulae used:** The Newton nth root theorem xn+1=xn−fn(x)f′n(x)

**Explanation:**

Consider f(x)=lnx+x−3

f(x)=lnx+x−3f′(x)=1x+1xn+1=xn−fn(x)f′n(x)xn+1=xn−lnxn+xn−31xn+1

For x>0, f′(x)=1x+1>0

The value of function at x=2, x=e are f(2)≈−0.307 and f(e)≈0.718 respectively

Therefore, f(x) is continuous on [2,e] and differentiable on (2,e)

Hence, use the formulae of Newton nth root theorem xn+1=xn−fn(x)f′n(x)

Therefore, various value of *x* given below

x1=2x2≈2.20457x3≈2.2079x4≈2.279

**Conclusion:** Hence the root of the expression lnx=3−x is 2.2079

#### Answer

Yes, we could prove that there is exactly one root of the given equation and it lies between (2,e).

**Given:** lnx=3−x

**Formulae used:** The intermediate value theorem

#### Explanation

Consider f(x)=lnx+x−3

f(x)=lnx+x−3f′(x)=1x+1

For x>0, f′(x)=1x+1>0

The value of function at x=2, x=e are f(2)≈−0.307 and f(e)≈0.718 respectively

Therefore, f(x) is continuous on [2,e] and differentiable on (2,e)

Hence, use the formulae of intermediate value theorem ∃c ∈(2,e), such that f(c)=0, it means that there is at least one root on this interval.

Thus, as seen from the above value f(2)<0 and f(e)>0

But f′(x)>0 for x∈(2,e), it means that the function is increasing on this interval and there can be only one root.

**Conclusion:** Hence there is exactly one root lie between (2,e)

**(** **b )**

**To find:** The root of the equation to four decimal places.

**Solution:** 2.2079

**Given:** The expression is lnx=3−x

**Formulae used:** The Newton nth root theorem xn+1=xn−fn(x)f′n(x)

Consider f(x)=lnx+x−3

f(x)=lnx+x−3f′(x)=1x+1xn+1=xn−fn(x)f′n(x)xn+1=xn−lnxn+xn−31xn+1

For x>0, f′(x)=1x+1>0

The value of function at x=2, x=e are f(2)≈−0.307 and f(e)≈0.718 respectively

Therefore, f(x) is continuous on [2,e] and differentiable on (2,e)

Hence, use the formulae of Newton nth root theorem xn+1=xn−fn(x)f′n(x)

Therefore, various value of *x* given below

x1=2x2≈2.20457x3≈2.2079x4≈2.279

**Conclusion:** Hence the root of the expression lnx=3−x is 2.2079