Question 1028474
<pre>

{{{e^x + e^(-x)}}}{{{""=""}}}{{{4}}}

A negative exponent means one over the base with the
positive exponent:

{{{e^x + 1/e^x}}}{{{""=""}}}{{{4}}}

Multiply through by {{{e^(x)}}}

{{{e^(2x) + 1}}}{{{""=""}}}{{{4e^x}}}

{{{e^(2x) - 4e^x + 1}}}{{{""=""}}}{{{0}}}

{{{(e^x)^2 - 4e^x + 1}}}{{{""=""}}}{{{0}}}

Let {{{u=e^x}}}

{{{u^2-4u+1}}}{{{""=""}}}{{{0}}}

 {{{u = (-b +- sqrt( b^2-4ac ))/(2a) }}}

 {{{u = (-(-4) +- sqrt( (-4)^2-4(1)(1) ))/(2(1)) }}}

 {{{u = (4 +- sqrt(16-4 ))/2 }}}

 {{{u = (4 +- sqrt(12))/2 }}}

 {{{u = (4 +- sqrt(4*3))/2 }}}

 {{{u = (4 +- 2sqrt(3))/2 }}}

 {{{u = (2(2 +- sqrt(3)))/2 }}}

 {{{u = (cross(2)(2 +- sqrt(3)))/cross(2) }}}

 {{{u = 2 +- sqrt(3) }}}

Since {{{u=e^x}}}

 {{{e^x = 2 +- sqrt(3) }}}

  {{{x = ln(2 +- sqrt(3)) }}}

{{{x="" +- 1.316957897}}}


The second way the other tutor did it, he mistakenly 
wrote sinh for cosh (hyperbolic functions) and got the 
wrong solution.  It should have been

cosh(x) = 2

which does gives the answer above:

{{{x="" +- 1.316957897}}}

Edwin</pre>