<p>I cannot figure this one out... can you help me???</p>
<p>Solve.</p>
<p>log (base 3) 2x^2 - log (base 3) (5x-9) = 1</p>
<p>any ideas??</p>
<p>Simple:
Combine the logs to get log (base 3) 2x^2/ 5x-9 = 1. Then raise both sides from a power of three to cancel out the log base 3. You have 2x^ 2/ 5x-9= 1, hence 2x^ 2= 5x-9 , and 0= 2x^2 + 5x - 9, use the quadratic formula.</p>
<p>You just need to know property of logs. The right side could be written as log (base 3) (2x^2)/(5x-9) =1. Then exponentiate both sides with 3, giving (2x^2)/(5x-9) =3^1=3. Now, you should be able to figure out the quadratic equation that remains, 2x^2-15x+27=0. You can factor to get (2x-9)(x-3)=0. Therefore, x is either 3 or 9/2. Plug both back into the original equation to determine which ones make sense.</p>
<p>Is that the kind of math people do @ UPENN? ("Ivy league14", pennsylvania, I just assumed you were from UPENN)</p>
<p>I have yet to begin college so correct me if I'm wrong. From what I gather, the big change between HS math and college math is that in college, instead of doing random problems (like in HS), you are applying it to some kind of situation.</p>
<p>Math in college, from my experience, is much more theoretical than math in high school.</p>
<p>Well doesn't it depend what major you're talking about?</p>
<p>Engineering, physics, chem, and applied math majors are applied (duh!)</p>
<p>A Math major is pure math, very theoretical.</p>
<p>I've always been curious about how math courses were different in college.</p>
<p>I hope that's not a math they study in college because logarithms are somewhere in school algebra.</p>
<p>no - log is in precalc</p>
<p>log (base 3) 2x^2 - log (base 3) (5x-9) = 1</p>
<p>log (base 3) (2x^2) / (5x-9) = 1
3 = (2x^2) / (5x-9)
0 = (2x^2) / (5x-9) - 3</p>
<p>solve</p>
<p>You didn't even solve it -_-"</p>
<p>Not to mention that someone else has alread posted the solution.</p>