<p>Can anybody help me on a few of the math problems on the practice test in the Official "Preparing for the ACT" little booklet/packet?</p>
<p>It's Math, Section 2:</p>
<p>-Question 42: What are the real solutions to the equation IxI^2 + 2IxI - 3 = 0
F. plus or minus 1
G. plus or minus 3
H. 1 and 3
J. -1 and -3
K. plus or minus 1 and plus or minus 3</p>
<p>The answer's F; but how?</p>
<p>-Question 54 (long one... could anybody with the packet possibly look at it and help?)</p>
<ol>
<li>
F... 1 + 2 - 3 = 0 Correct
G... 9 + 6 - 3 =/= 0 Incorrect
H... 3 doesn't work, so this answer is Incorrect as well
J... -3 and 3 give you the same result, so this also doesn't work
K... Anything with 3 won't give you a correct result</li>
</ol>
<p>Therefore, the answer has to be F.</p>
<p>I'm too lazy to check 54 cuz i'm lazy, maybe later, but i hope that helped</p>
<p>i thought those were just parentheses...i hate doing math problems on this forum...you cant do exponents without ^ and all that other annoying stuff. thanks for pointing that out though.</p>
<p>Number 60 in that same packet in 2nd section (Math)</p>
<ol>
<li>What is the real value of x in the equation
log(subscript:2)24 - log(subscript:2)3 = log(subscript:5)x?</li>
</ol>
<p>I sort of forgot the rules of logarithms so can somebody remind me? I looked up that log(subscript:a)x = b is equivalent to x = a^b.... but im not sure how to answer this problem, btw in which the answer is J. 125</p>
<p>when subtracting logs with the same bases, you can change it into one log by dividing the exponents. in this example, log (base2) 24 - log(base2)3 is the same as log(base2) 8---->24/3. so now we have log(base2) 8=log(base5)x. well log(base2)8 =3, so the final equation is 5^3=x. x=125.</p>