Question 719905
{{{e^x-e^(-x)=10}}}
Rewriting {{{e^(-x)}}} as {{{1/e^x}}} will help us see what to do:
{{{e^x-1/e^x=10}}}
First, we'll eliminate the fraction by multiplying both sides by {{{e^x}}}:
{{{e^x*e^x-e^x*(1/e^x) = e^x*10}}}
which simplifies to:
{{{e^(2x)-1 = 10e^x}}}
Next we'll get a zero on the right side by subtracting {{{10e^x}}}:
{{{e^(2x)-10e^x-1 = 0}}}
This does not factor, unfortunately.<br>
This next part is probably the hardest part. The exponent of {{{e^(2x)}}} is exactly twice the exponent of {{{10e^x}}}. This makes this equation an equation in what is called "quadratic form". Quadratic form equations can be solved in much the same way as regular quadratic equations. Using a temprary variable can help you see the "quadratic-ness" of our equation. Set it equal to the base and the smaller exponent:
Let {{{q = e^x}}}. Then {{{q^2 = (e^x)^2 = e^(2x)}}}. Substituting these into our equation we get:
{{{q^2-10q-1=0}}}
This is clearly an quadratic equation. As noted earlier, this will not factor. But we can use the quadratic formula:
{{{q = (-(-10) +- sqrt((-10)^2-4(1)(-1)))/2(1)}}}
Simplifying...
{{{q = (-(-10) +- sqrt(100-4(1)(-1)))/2(1)}}}
{{{q = (-(-10) +- sqrt(100+4))/2(1)}}}
{{{q = (-(-10) +- sqrt(104))/2(1)}}}
{{{q = (10 +- sqrt(104))/2}}}
{{{q = (10 +- sqrt(4*26))/2}}}
{{{q = (10 +- sqrt(4)*sqrt(26))/2}}}
{{{q = (10 +- 2*sqrt(26))/2}}}
{{{q = (2(5 +- sqrt(26)))/2}}}
{{{q = (cross(2)(5 +- sqrt(26)))/cross(2)}}}
{{{q = 5 +- sqrt(26)}}}
which is short for:
{{{q = 5 + sqrt(26)}}} or {{{q = 5 - sqrt(26)}}}<br>
Of course we are not interested in solutions for q. We are interested in solutions for x. So now we substitute back for the q:
{{{e^x = 5 + sqrt(26)}}} or {{{e^x = 5 - sqrt(26)}}}
We have a little more work to do to solve for x. Since {{{sqrt(26) > 5}}} the second equation says that {{{e^x}}} is a negative number. But it is impossible for e to <i>any</i> power to be a negative number. So that equation is impossible. There will be no solutions for x from that equation.<br>
So our only solution(s) will come from the first equation. Finding the ln of each side:
{{{ln(e^x) = ln(5 + sqrt(26))}}}
On the left side we use a property of logs, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent out in front:
{{{x*ln(e) = ln(5 + sqrt(26))}}}
Since ln(e) = 1 the left side becomes:
{{{x = ln(5 + sqrt(26))}}}
This is an exact expression for the solution to your equation. If you want/need a decimal approximation, get out your calculator.<br>
P.S. After you have done a few of these quadratic form equations, you will not need a temporary variable. You will start to see how to go directly from:
{{{e^(2x)-10e^x-1 = 0}}}
to
{{{e^x = (-(-10) +- sqrt((-10)^2-4(1)(-1)))/2(1)}}}
etc.