How do you solve for x, ln ( (x^2) +1)=cos x??

Question

How do you solve for x, ln ( (x^2) +1)=cos x??
I need to find x for the limits of integration for integral of cosx- ln (x^2 +1) dx, to find the area, volume, and arc length, of the region they bounded. My calculator gives an estimate, as x= + or - .9158576564, but id like an exact value. Thanks alot for the help. Its Greatly apperciated.

Answer 1

first you need to exponetiate each side so you have

e^ In((x^2)+1)= e^cosx
then by canceling e and ln you have
x^2 + 1= e^cosx
substract one
x^2=e^cosx-1
take the square root of both sides
x= +/- square root of e^cosx -1
i think this way might also work:
ln ( (x^2) +1)=cos x
take the inverse cos of both sides
cos^ -1(ln((x^2) + 1)= x

Answer 2

This is a transcendent equation which cant be solved by algebraic
,means.The solution is surely a real number with an infinite number of digits
Just a question Why would you think that sqrt2 is an exact solution?
Your calculator would give you 1.4142......
Ten digits are normally more than enough
My calculator
gives me 0.915857659124 .As the expression is even - would be solution

Answer 3

i visit think of that log refers to base 10. the main thought is to apply the replace of base theorem interior the 1st step. Then we are able to unravel for ln(x) and then resolve for x. a million + log(x) = ln(x) a million + [ln(x)/ln(10)] = ln(x), from the replace of base theorem ln(10) + ln(x) = ln(x) ln(10), from multiplying the two factors with the aid of ln(10) ln(10) = ln(x) ln(10) - ln(x), from subtracting ln(x) from the two factors ln(10) = ln(x) * [ln(10) - a million], from factoring out ln(x) on the excellent suited-hand facet ln(x) = ln(10) / [ln(10) - a million] x = e^{ln(10) / [ln(10) - a million]} x = {e^[ln(10)]} ^ {a million/[ln(10) - a million]} x = 10 ^ {a million/[ln(10) - a million]} or approximately 5.8574 Lord bless you at present!