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ok, let's say you want to find log 12 (i'm assuming it's base 10?) we can write this as \[\log(1.2*10^1)=\log(1.2)+\log(10^1)=\log(1.2)+1\] that first step is an important rule with logarithms. Now, if you have a table, let's find 1.2 on it. It's probably broken down into sections based on the first digit or two, and then you have to read across. There's probably a 1.2 row, so you look in that row and go over to the 0 column; this means you're finding the log of 1.20, which is clearly the same as 1.2 since i'm not sure what your table is like, try it and tell me if that's how it's set up. If not, describe it to me (or link)
thats how its set up
sweet, then all you do is take that number and add 1 to it. similarly, for log 0.038, you put it in scientific notation, break it up like I did, and then find the value of 3.8 in the log table, then add.
give me one sec
so for 1.2 it would = 0.0792
so do i move the decimal once or twice because you broke 0.038 into 3.8 and 12 into 1.2
you move it as many times as necessary to make it so that there is one digit to the left of the decimal point. Then, the exponent on the ten becomes the number of places you moved it. If you move it to the left, the exponent, is positive, and if it's to the right, it's negative.
for example, 34500 is 3.45*10^4 and .0065 is 6.5*10^(-3)
what about this one 21,100
well, how many places would you have to move the decimal point to make it so that exactly 1 digit is on the left side of it?
how do i find that on the table
that's 4, not 1... you go to the 2.1 row, then over to the 1 column
owww your right forgot about that
but since its 2.11 i find 2.1 the go over to one right
yeah buddy im getting the hang of this
thing is im in the ACE program and the book doesnt really explain it as a one on one so its much harder
well, i'm glad to be of help
im glad to be helped lol