<p>e^x + e^(-x) = 3</p>
<p>Solve for x.</p>
<p>So I just came across this problem, but I can't seem to isolate x. I tried adding the left side by first simplifying e^(-x) to 1/(e^x), and then adding the two terms, but that doesn't get me anywhere. </p>
<p>Is there a general model for solving transcendental equations? </p>
<p>Thanks for your input.</p>
<p>use e^x=y</p>
<p>it reduces to a quadratic</p>
<p>but how can you set it equal to y?</p>
<p>(y^2+1)/y = 3 </p>
<p>thats what you get. What I don't understand is how you would solve for y</p>
<p>oh never mind. i got it =) didn't realize i should multiply the y to the other side =P.</p>
<p>Thanks shash</p>
<p>I believe you could also multiply by e^x.</p>
<p>e^x(e^x + e^-x) = 3(e^x)</p>
<p>e^2x + 1 = 3e^x</p>
<p>e^2x - 3e^x +1 = 0</p>
<p>Substitute x for e^x.</p>
<p>x^2 - 3x + 1 = 0</p>
<p>Quadratic Formula, solve for x, set that equal to e^x and you're done.</p>
<p>I'm not sure if this method is easier, but it's what I have always done with this type of problem.</p>
<p>Its the same thing, but you substitue a varible for e^x later rather than sooner</p>
<p>Also, it's always safer to substitute a different variable than x - on a test you can end up finding x, but not the final one.</p>