<p>When you subtract logs from each other it's the same thing as dividing them.</p>
<p>ex.</p>
<p>log(a-1) - log(b-1) = log(a-1)/(log(b-1)</p>
<p>Also, on previous ACTs and practice tests I usually have 20-25 minutes left when I'm done. I literally went up to the last second on this test. Of course there is always going to be a test day anxiety that will slow the process down, but this particular test was quite hard. </p>
<p>ugh i failed miserably...so annoyed. got a 29 last time but i guessed on 10 cuz i ran out of time and was expecting a lot worse score. i guessed on like 3 this time, but i think i made a ton of careless mistakes on the others:(</p>
<p>For the majority of you that have taken the ACT before, has this been the hardest math section you've ever seen? Just wondering because it was my first time today and I went crazy. I usually finish at least with 5-10min left on practice tests and I didn't even get to finish today.</p>
<p>What I did was just set my calc to radians set a degree for x... i made it 65 or something and i tested all of the answers and got 2cscx (or it was just that none of the others worked...)</p>
<p>i thought the banquet one was as simple as it looked. i just skimmed through the stuff cause i didnt have time... but i think it asked how many people would be at the banquet and it gave numbers for all three sports</p>
<p>i think you just add those up because im guessing that when it said the people who did 2 and 3 sports were already counted in participating in the one sport</p>
<p>the banquet one was 55, cuz you have to subtract the people who were counted twice. and yeah, i thought this time the math section was kinda hard...</p>
<p>I got 70 for that question, however ppl don't seem to be getting that.
On the last question, I actually thought it was one of the easiest questions on the test. All you had to do was plug in a value for x (degree) value..plug it in..and see which answer choice was right. On84..I idnt know how to input the last choice but still knew it was correct cuz nothing else was right.</p>
<p>jckund, did radians work for you? I think I got your answer and used your same method. Actually I think I might have had it on radian but I think you only want to do that if you're graphing it, not if you're doing regular calculations. I thought I used degree though I really can't remember.</p>