SOLUTION: Find the exact value of x for which 4cosh(x) - e^-(x) = 3 Find the exact value of cosh^-¹(3) - sinh^-1(2)

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Question 1196149: Find the exact value of x for which 4cosh(x) - e^-(x) = 3
Find the exact value of cosh^-¹(3) - sinh^-1(2)
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
4cosh%28x%29+-+e%5E%28-x%29+=+3............................use identity cosh%28x%29+=e%5E%28-x%29%2F2+%2B+e%5Ex%2F2
4%28e%5E%28-x%29%2F2+%2B+e%5Ex%2F2%29-+e%5E%28-x%29+=+3
2e%5E%28-x%29+%2B+2+e%5Ex-e%5E%28-x%29+=3
+e%5E%28-x%29+%2B+2e%5Ex+=3
e%5E%28-x%29+%282+e%5E%282+x%29+%2B+1%29+=3
2e%5E%282+x%29+%2B+1=3%2Fe%5E%28-x%29+
2e%5E%282x%29+%2B+1=3e%5Ex
2e%5E%282x%29+-3e%5Ex%2B1=0

Rewrite the equation with: e%5Ex=u
2u%5E2-3u=-1
2u%5E2-3u%2B1=0
%282u+-+1%29%28u+-+1%29=0
solutions for u:
if %282+u+-+1%29+=0+=>+2u=1 =>u=1%2F2
if +%28u+-+1%29=0 =>u=1

Substitute back u=e%5Ex, solve for+x
u=1%2F2
e%5Ex=1%2F2+....take natural log of both sides
ln%28e%5Ex%29=ln%281%2F2%29
+x%2Aln%28e%29=ln%281%2F2%29
x%2A1=ln%281%29-ln%282%29
x=0-ln%282%29
x+=+-ln%282%29
or
u=1
e%5Ex=1
ln%28e%5Ex%29=ln%281%29
x%2Aln%28e%29=0
x%2A1=0
x=0
verify solutions:
4cosh%28-ln%282%29%29+-+e%5E%28-%28-ln%282%29%29%29+=+3
4%285%2F4%29+-+2%29+=+3
5+-+2%29+=+3
3%29+=+3 => true
or

4cosh%280%29+-+e%5E%28-0%29+=+3
4%281%29+-+1+=+3
3=+3



cosh%5E-1%283%29+-+sinh%5E-1%282%29

Use the following identities:
sinh%5E-1%28x%29=ln%28sqrt%281%2Bx%5E2%29%2Bx%29
cosh%5E-1%28x%29=ln%28sqrt%28-1%2Bx%5E2%29%2Bx%29

so we have

ln%28sqrt%28-1%2B3%5E2%29%2B3%29-+ln%28sqrt%281%2B2%5E2%29%2B2%29
=ln%282sqrt%282%29+%2B+3%29-+ln+%28sqrt%285%29+%2B+2%29
=ln%28%282sqrt%282%29+%2B+3%29%2F%28sqrt%285%29+%2B+2%29%29+-> exact solution





Answer by ikleyn(52782) About Me  (Show Source):
You can put this solution on YOUR website!
.
(a) Find the exact value of x for which 4cosh(x) - e^-(x) = 3

(b) Find the exact value of cosh^-¹(3) - sinh^-1(2)
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            There are two problems in the post, which is prohibited by the rules of the forum.

            So,  I will solve part  (a)  ONLY,  in order for do not transform my response into a mess. 


By the definition,  cosh(x) = %28e%5Ex%2Be%5E%28-x%29%29%2F2.


So, I will introduce new variable  t = e%5Ex.  


Then the equation takes the form

    4%2A%28%28t%2B1%2Ft%29%2F2%29 - 1%2Ft = 3.


Simplify it step by step

    2%2A%28t%2B1%2Ft%29 - 1%2Ft = 3

    2t + 1%2Ft = 3

    2t^2 + 1 = 3t

    2t^2 - 3t + 1 = 0.


Apply the quadratic formula.  It will give two roots:  t = 1  and  t = 1%2F2.


If t = 1,  then  e%5Ex = 1  and  x = 0.

If t = 1%2F2,  then  e%5Ex = 1%2F2  implies  x = ln%281%2F2%29 = -ln(2).


Thus part (a) has two solutions  x = 0  and  x = -ln(2).    ANSWER

Part (a) is just solved.

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