<p>I am sure many of you on here are a lot smarter than me...so maybe this is the right place to ask (I don't have a book to look any of this up). </p>
<p>e^(1+lnx)</p>
<p>e^(3lnx)</p>
<p>3sin^2x=cos^2x (x is between 0 and 360)</p>
<p>sin2x= sinx (x is between 0 and 360)</p>
<p>thanks alot for any help...the packet was like 8 pages long so at least I didn't forget too much :) .</p>
<p>Hi there,</p>
<p>Here are the solutions that I came up with, but it is important that you understand how it all works:</p>
<p>e^(1+lnx) = e^1 * e^(ln x). e^1 is just e. e^(ln x) is x. Notice that the base of the natural log is e, e ^ (ln of whatever) is just whatever. Final answer is ex</p>
<p>e^(3lnx) = e^(ln x^3). Notice that the 3 goes back up as an exponent. This simplifies to x^3</p>
<p>3sin^2x=cos^2x (x is between 0 and 360). cos^2(x) is 1-sin^2(x). So we have 3sin^2(x) = 1 - sin^2(x). 4sin^2(x) = 1. sin^2(x) = 1/4. You should be able to get the rest.</p>
<p>sin2x= sinx (x is between 0 and 360) sin2x = 2sinxcosx. So we now have 2sinxcosx=sinx. 2sinxcosx - sinx = 0. sinx(2cosx - 1) = 0. This means sinx is 0, or 2cosx - 1 is 0. You can figure out the rest</p>