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Popular Trigonometry >

cosh(x-1)=2

  • Pre Algebra
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Solution

cosh(x−1)=2

Solution

x=ln(e(2+3​)),x=ln(e(2−3​))
+1
Degrees
x=132.75190…∘,x=−18.16034…∘
Solution steps
cosh(x−1)=2
Rewrite using trig identities
cosh(x−1)=2
Use the Hyperbolic identity: cosh(x)=2ex+e−x​2ex−1+e−(x−1)​=2
2ex−1+e−(x−1)​=2
2ex−1+e−(x−1)​=2:x=ln(e(2+3​)),x=ln(e(2−3​))
2ex−1+e−(x−1)​=2
Multiply both sides by 22ex−1+e−(x−1)​⋅2=2⋅2
Simplifyex−1+e−(x−1)=4
Apply exponent rules
ex−1+e−(x−1)=4
Apply exponent rule: ab+c=abacex−1=exe−1,e−(x−1)=e−1xe1exe−1+e−1⋅xe1=4
Apply exponent rule: abc=(ab)ce−1x=(ex)−1exe−1+(ex)−1e1=4
exe−1+(ex)−1e1=4
Rewrite the equation with ex=uue−1+(u)−1e1=4
Solve ue−1+u−1e1=4:u=e(2+3​),u=e(2−3​)
ue−1+u−1e1=4
Refinee1​u+ue​=4
Multiply by LCM
e1​u+ue​=4
Find Least Common Multiplier of e,u:eu
e,u
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear either in e or u=eu
Multiply by LCM=eue1​ueu+ue​eu=4eu
Simplify
e1​ueu+ue​eu=4eu
Simplify e1​ueu:u2
e1​ueu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=e1​eu1+1
Add the numbers: 1+1=2=e1​eu2
Multiply fractions: a⋅cb​=ca⋅b​=e1e​u2
Cancel the common factor: e=u2⋅1
Multiply: u2⋅1=u2=u2
Simplify ue​eu:e2
ue​eu
Multiply fractions: a⋅cb​=ca⋅b​=ueeu​
Cancel the common factor: u=ee
Apply exponent rule: ab⋅ac=ab+cee=e1+1=e1+1
Add the numbers: 1+1=2=e2
u2+e2=4eu
u2+e2=4eu
u2+e2=4eu
Solve u2+e2=4eu:u=e(2+3​),u=e(2−3​)
u2+e2=4eu
Move 4euto the left side
u2+e2=4eu
Subtract 4eu from both sidesu2+e2−4eu=4eu−4eu
Simplifyu2+e2−4eu=0
u2+e2−4eu=0
Write in the standard form ax2+bx+c=0u2−4eu+e2=0
Solve with the quadratic formula
u2−4eu+e2=0
Quadratic Equation Formula:
For a=1,b=−4e,c=e2u1,2​=2⋅1−(−4e)±(−4e)2−4⋅1⋅e2​​
u1,2​=2⋅1−(−4e)±(−4e)2−4⋅1⋅e2​​
(−4e)2−4⋅1⋅e2​=23​e
(−4e)2−4⋅1⋅e2​
(−4e)2=42e2
(−4e)2
Apply exponent rule: (−a)n=an,if n is even(−4e)2=(4e)2=(4e)2
Apply exponent rule: (a⋅b)n=anbn=42e2
4⋅1⋅e2=4e2
4⋅1⋅e2
Multiply the numbers: 4⋅1=4=4e2
=42e2−4e2​
42=16=16e2−4e2​
Add similar elements: 16e2−4e2=12e2=12e2​
Apply radical rule: assuming a≥0,b≥0=12​e2​
12​=23​
12​
Prime factorization of 12:22⋅3
12
12divides by 212=6⋅2=2⋅6
6divides by 26=3⋅2=2⋅2⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅2⋅3
=22⋅3
=22⋅3​
Apply radical rule: =3​22​
Apply radical rule: 22​=2=23​
=23​e2​
Apply radical rule: assuming a≥0e2​=e=23​e
u1,2​=2⋅1−(−4e)±23​e​
Separate the solutionsu1​=2⋅1−(−4e)+23​e​,u2​=2⋅1−(−4e)−23​e​
u=2⋅1−(−4e)+23​e​:e(2+3​)
2⋅1−(−4e)+23​e​
Apply rule −(−a)=a=2⋅14e+23​e​
Multiply the numbers: 2⋅1=2=24e+23​e​
Factor 4e+23​e:2e(2+3​)
4e+23​e
Rewrite as=2⋅2e+2e3​
Factor out common term 2e=2e(2+3​)
=22e(2+3​)​
Divide the numbers: 22​=1=e(2+3​)
u=2⋅1−(−4e)−23​e​:e(2−3​)
2⋅1−(−4e)−23​e​
Apply rule −(−a)=a=2⋅14e−23​e​
Multiply the numbers: 2⋅1=2=24e−23​e​
Factor 4e−23​e:2e(2−3​)
4e−23​e
Rewrite as=2⋅2e−2e3​
Factor out common term 2e=2e(2−3​)
=22e(2−3​)​
Divide the numbers: 22​=1=e(2−3​)
The solutions to the quadratic equation are:u=e(2+3​),u=e(2−3​)
u=e(2+3​),u=e(2−3​)
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of ue−1+u−1e1 and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=e(2+3​),u=e(2−3​)
u=e(2+3​),u=e(2−3​)
Substitute back u=ex,solve for x
Solve ex=e(2+3​):x=ln(e(2+3​))
ex=e(2+3​)
Apply exponent rules
ex=e(2+3​)
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(e(2+3​))
Apply log rule: ln(ea)=aln(ex)=xx=ln(e(2+3​))
x=ln(e(2+3​))
Solve ex=e(2−3​):x=ln(e(2−3​))
ex=e(2−3​)
Apply exponent rules
ex=e(2−3​)
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(e(2−3​))
Apply log rule: ln(ea)=aln(ex)=xx=ln(e(2−3​))
x=ln(e(2−3​))
x=ln(e(2+3​)),x=ln(e(2−3​))
x=ln(e(2+3​)),x=ln(e(2−3​))

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Frequently Asked Questions (FAQ)

  • What is the general solution for cosh(x-1)=2 ?

    The general solution for cosh(x-1)=2 is x=ln(e(2+sqrt(3))),x=ln(e(2-sqrt(3)))
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