any help will be appreciated....thank you
best i got out of this was x=ln{1+ or - 2^(1/2)}i ...but unsure if that is correct or even a valid expression...thank you for any help ...peace
Initial equation
$\displaystyle cosh(x)=i$
Take inverse
$\displaystyle x=arcosh(i)$
Identity
$\displaystyle x=ln(i+\sqrt{i-1}\sqrt{i+1})$
Multiply Square Roots
$\displaystyle x=ln(i+\sqrt{(i-1)(i+1)})$
FOIL
$\displaystyle x=ln(i+\sqrt{i^2-1})$
Simplify
$\displaystyle x=ln(i+\sqrt{-2})$
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VERIFICATION:
Initial Equation
$\displaystyle cosh(x)=i$
Identity
$\displaystyle \frac{e^x}{2} + \frac{e^{-x}}{2}=i$
Rewrite
$\displaystyle \frac{e^x}{2} + \frac{1}{2e^x}=i$
Substitute in our Answer
$\displaystyle \frac{e^{ln(i+\sqrt{-2})}}{2} + \frac{1}{2e^{ln(i+\sqrt{-2})}}=i$
Simplify
$\displaystyle \frac{i+\sqrt{-2}}{2} + \frac{1}{2(i+\sqrt{-2})}=i$
Multiply by 2 and subtract i from both sides
$\displaystyle \sqrt{-2} + \frac{1}{i+\sqrt{-2}}=i$
Common denominator
$\displaystyle \frac{i\sqrt{-2}+\sqrt{-2}\sqrt{-2}}{i+\sqrt{-2}} + \frac{1}{i+\sqrt{-2}}=i$
Simplify
$\displaystyle \frac{i^2\sqrt{2}-\sqrt{2}^2+1}{i+\sqrt{-2}}=i$
Simplify
$\displaystyle \frac{-\sqrt{2}-1}{i+\sqrt{-2}}=i$
Multiply by the Denominator
$\displaystyle -\sqrt{2}-1=i(i+\sqrt{-2})$
Distribute
$\displaystyle -\sqrt{2}-1=i^2+i\sqrt{-2}$
Simplify
$\displaystyle -\sqrt{2}-1=i^2+i^2\sqrt{2}$
Simplify
$\displaystyle -\sqrt{2}-1=-1-\sqrt{2}$
Both sides are equal, thus the answer is correct.