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Homework Help

Discussion in 'Offtopic' started by Masteraj12, May 6, 2017.

  1. Masteraj12

    Masteraj12 (☞゚ヮ゚)☞Sr. Moderatorororor☜(゚ヮ゚☜) Sr. Mod

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    So let's start by giving you the answer. x, in this problem, equals 0

    .5^x =1 is a basic log property we can derive. If you have anything to the power of whatever, it produces a number. However, 0 is a special case. Anything to the power of 0 is going to always equal 1. 2^0=1, x^0=1, etc. Knowing this, we know where we want to go to, now we just need to get there.

    In log problems, we generally take the log of both sides to isolate the x. If we take the log of both sides of .5^x=1, then we end up with x*log.5 = log1. Whenever you take a log of both sides you can bring down the exponent and multiply it by the log. Example, if we take the log of 2^x = 4, we do logs on both sides and bring down the exponents, getting x*log2 = log4. We can then solve for that with our calculator to get that x = 2 (and then again, we can just visually see that if we raise 2 to the power of 2, it equals 4).

    I mentioned earlier that 0 is special, and so is 1. Basically, when we are taking the log of a number, say 4, if we do not specify a base of the log (so we say Log Base 3, or Log Base 4), then we assume the log is ALWAYS based 10. So say we take log4. What we are essentially trying to find is what exponent makes the base of the log that number inside. In this example, we are trying to find what exponent makes 10 turn into 4. You can plug that into your calculator and find out it works, but thats the core concept. Now in our problem, we have x*log.5 = log1. Trying to do this w/o a calculator is easy, because what we can do is look at the right side which is log1. What we want here is to find out what exponent makes 10 go to 1. WAIT! That's 0, as anything to the power of 0 is 1. So now we can make it become x*log.5 = 0. We then move to isolate x by dividing log.5 over, and we find that x=0. And it makes sense, raising .5^0 will make it equal 1. Now you don't have to do these steps every time, but this is just building from what logs do to proving the general rule.

    hmu on teamspeak if you need some more help
     
  2. MineButcher

    MineButcher Well-Known Member

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    isn't anything to the power of 0, 1?

    you are making me wish i never read this, my head hurts a little bit
     
  3. Masteraj12

    Masteraj12 (☞゚ヮ゚)☞Sr. Moderatorororor☜(゚ヮ゚☜) Sr. Mod

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    I was just giving a reasoning for when you solve it you can just skip the math and say it's 0. Legit second sentence I saw that rule...
     
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  4. Masteraj12

    Masteraj12 (☞゚ヮ゚)☞Sr. Moderatorororor☜(゚ヮ゚☜) Sr. Mod

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    Heyo bumping this thread because March-May are the heaviest months for school, don't want anyone overstressing.
     
  5. Draconium

    Draconium Well-Known Member

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    Can someone explain the Factor Theorum? I tried searching it up and I’m still a bit confused by it. O and it’s Alg. 2
     
  6. Masteraj12

    Masteraj12 (☞゚ヮ゚)☞Sr. Moderatorororor☜(゚ヮ゚☜) Sr. Mod

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    @Draconium So basically if you have a function, you can find out some of it's roots by dividing out a certain answer.
    We have the function x^2 - 2x + 1 right. Now we can simply take the two roots by factoring out, x-1 times x-1 right. Using the factor theorem, you can do x^2 -2x +1 divided by x-1 to find out if thats a root. If when you divide, the remainder is 0, then that function you divide is a root.

    EXAMPLE: Use the Factor Theorem to find out if x-1 is a root of 2x^4 - 3x^3+1.
    Try doing this on your own by using either long division or using synthetic division and check the spoiler for the answer/steps.
    So I'm going to use synthetic division because it makes it EASY AS HELL
    If you don't know synthetic division, just divide (2x^4 - 3x^3 +1) by x-1.
    upload_2018-3-2_13-29-57.png
    Looking at this, we find out that x-1 IS a root because it divides cleanly and we have a remainder of 0.
     

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