Answer: D) x=ln√t+1/1-t
Step-by-step explanation:
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Therefore value of x = [tex]\rm ln \sqrt{ \dfrac{1+t}{1-t}\\[/tex] , Option D is the correct answer.
Equations are mathematical expressions equated by equal sign , it has a wide applications in various field of mathematics , Quadratic Equations , Mathematical equations etc.
It is given in the question that
e^x-e^-x/e^x+e^-x=t
[tex]\rm \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = t\\\\\\\rm {e^x - e^{-x}} = t ({e^x + e^{-x}})\\\\\\e^x -t e^x = te^{-x} + e^{-x}\\\\\\(1-t)e^x = (1+t)e^{-x}\\\\\\e^{2x} = \dfrac{1+t}{1-t}\\\\\\Applying \; log \; on \;both\;sides\\\\\\\\2x = ln \dfrac{1+t}{1-t}\\\\\\x = \dfrac{1}{2} ln \dfrac{1+t}{1-t}\\\\\\\\x = ln \sqrt{ \dfrac{1+t}{1-t}\\[/tex]
Therefore value of x = [tex]\rm ln \sqrt{ \dfrac{1+t}{1-t}\\[/tex] , Option D is the correct answer.
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