Question 689684
{{{2e^x=5-e^(-x)}}}
I am going to multiply each side by {{{e^x}}}. You will better understand why if I first rewrite {{{e^(-x)}}} as {{{1/e^x}}}:
{{{2e^x=5-1/e^x}}}
Multiplying:
{{{e^x*(2e^x)=e^x*(5-1/e^x)}}}
we get:
{{{2e^(2x)=5e^x -1}}}
Since the exponent of {{{e^(2x)}}} is exactly twice the exponent of {{{e^x}}} this equation is in what is called quadratic form. To see this better I am going to use a temporary variable. Let {{{q = e^x}}}. Then {{{q^2 = (e^x)^2 = e^(2x)}}}. Substituting these in we get:
{{{2q^2 = 5q-1}}}
The equation is obviously quadratic. We first want one side to be zero. Subtracting the entire right side from each side we get:
{{{2q^2-5q +1=0}}}
Next we factor or use the Quadratic Formula. This will not factor so we must use the formula:
{{{q = (-(-5) +- sqrt((-5)^2-4(2)(1)))/2(2)}}}
which simplifies as follows:
{{{q = (-(-5) +- sqrt(25-4(2)(1)))/2(2)}}}
{{{q = (-(-5) +- sqrt(25-8))/2(2)}}}
{{{q = (-(-5) +- sqrt(17))/2(2)}}}
{{{q = (5 +- sqrt(17))/4}}}
which is short for:
{{{q = (5 + sqrt(17))/4}}} or {{{q = (5 - sqrt(17))/4}}}
We now have solutions for q. But we are looking for solutions for x. Now we substitute back in for q. (It was temporary, remember?)
{{{e^x = (5 + sqrt(17))/4}}} or {{{e^x = (5 - sqrt(17))/4}}}
Now we solve these equations for x. (NOTE: Since {{{sqrt(17)}}} is less than 5, the right side of both equations will be positive. If we had gotten zero or a negative for the right side of either equation there would be no solution for that equation since {{{e^x}}} can <b>never</b> be zero or negative.)<br>
We solve for x by finding the ln of each side:
{{{ln(e^x) = ln((5 + sqrt(17))/4)}}} or {{{ln(e^x) = ln((5 - sqrt(17))/4)}}}
On the left sides 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(17))/4)}}} or {{{x*ln(e) = ln((5 - sqrt(17))/4)}}}
And since ln(e) = 1 these become:
{{{x = ln((5 + sqrt(17))/4)}}} or {{{x = ln((5 - sqrt(17))/4)}}}
These are exact expressions for the solutions to your equation. If you want/need decimal approximations get out your calculator.<br>
P.S. Once you have done a few of these quadratic form equations, you will not need to use a temporary variable. You will see how to go from
{{{2e^(2x)=5e^x -1}}}
to
{{{2e^(2x)-5e^x +1=0}}}
to
{{{e^x = (-(-5) +- sqrt((-5)^2-4(2)(1)))/2(2)}}}
etc.