<p><a href="http://www.actstudent.org/pdf/preparing.pdf%5B/url%5D">http://www.actstudent.org/pdf/preparing.pdf</a></p>
<p>number 6o on the the math section...the very last math question...i dont get it
log2^24 - log2^3 = log5^x</p>
<p>WHAT IS X!?!?!</p>
<p>There's probably an easier way but what I did was:</p>
<p>log2(24) - log2(3) = log(5)x
log2(8) = log(5)x
log 8/log 2 = log x/log 5
log x = (log 8 * log 5)/log 2
x = 10^[(log 8 * log 5)/log 2]
x = 125</p>
<p>Or you can do:</p>
<p>log2(24) - log2(3) = log(5)x
log2(8) = log(5)x
x = 5^[log2(8)]
x = 5^(log 8/log 2)
x = 125</p>
<p>I hope you understood what I did :X.</p>
<p>o when you subtact logarithms its like division sortaa...</p>
<p>yes.. Its a log rule... Google it..</p>
<p>or the long way
log2^24 - log2^3 = log5^x</p>
<p>log2^24/log2^3= log2^8
when you plug into your calc. log2^8= 3
3=log5^X<br>
(you could stop here if you know that x=125) or keep going;
3=(log10^X/log10^5) (Change of Base)
3(log10^5)=log10^X
2.097 (roughly)= log10^X
exponents get rid of logs
10^2.097=10^log10^X
10^2.097=X
125=X</p>
<p>Of course Rwave just showed that....</p>
<p><_></p>
<p>i figured change of base wud be easiest..
log24/log2 - log3/log2 = log5(x)
log24/log2 - log3/log2 (in your calc.) = 3
so log5(x) = 3 and using the rules of log,
5^3 =x = 125</p>
<p>For some reason, my high school math teacher decided it would be wise to skip the lesson on logs, and I didn't do quite enough studying of them apparently before the ACT test. It was the last question on my test - different than the above, but still on logs. I pretty much had to whip out the brute force method and start plugging answer choices in for x until I found one that made the equation roughly equal.</p>
<p>That's the good thing about the ACT Math; there's almost always multiple ways to get the same answer.</p>
<p>its the log rule combined with algebra</p>
<p>The TI-83+ equation solver (Math, Up, Enter) can solve this for you if you remember your change-of-base formula. </p>
<p>Calculators shouldn't be allowed on this test. :(</p>
<p>(you could stop here if you know that x=125) or keep going;
3=(log10^X/log10^5) (Change of Base)
3(log10^5)=log10^X
2.097 (roughly)= log10^X
exponents get rid of logs
10^2.097=10^log10^X
10^2.097=X
125=X </p>
<p>all this is rubish.</p>
<p>you all seem not to know the basic log rule, which is
loga^x=y , a^y=x</p>
<p>the easiest way to solve this quesiton is:</p>
<p>log2^24 - log2^3 = log5^x
log2^(24/3)=log5^x
log2^8=log5^x
now solve log2^8 which is simply 2^y=8, y=3. Now if you cannot see that for 2 to become 8 you have to cube it; you can put Log8/log2 (change of base rule) in your calculator (but that is just a waste of time.)
so now u have 3=log5^x, which means that 5^3=x, so x=125</p>