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Math Help - I'm having trouble with these logs...not sure how to get started.

  1. #1
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    I'm having trouble with these logs...not sure how to get started.

    log4^(x+7)-log4^(x-3)

    I may have figured out this one...

    (x+7)-(x-3)
    x=-(-7-3)
    x=10
    Is this right???


    and

    logc^x+logc^y

    also this one, but it's different from the first two i think...any help getting started would be terrific!

    log2^(x+1)+log2^(x-5)=4

    I may have figured out the last problem...

    log2^(x+1)+log2^(x-5)=4
    log2x+log2(1)=log2+x+log2-5=4
    x+1+x-5=4
    x-4=4
    x=0
    ??? Is this right ???
    Last edited by Mhockey02; March 22nd 2010 at 06:24 PM.
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  2. #2
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    Log4(x+7) - log4(x-3) = Log4((x+7)/(x-3)), i don't know how far you have to go with this problem, however you can simplify it to

    Log4( 1 + 10/(x+3))

    Logc(x) + logc(y) = Logc(xy), again im not quite sure what they want you to do with this. If it is Logc(xy) = 0 then it can be simplified to xy = 1... so y = 1/x??

    Log2(x+1) + log2(x-5) = 4;
    Log2((x+1)(x-5)) = 4;
    Log2(x^2 - 4x - 5) = 4;
    x^2 - 4x - 5 = 2^4;
    x^2 - 4x - 5 = 16;
    x^2 - 4x - 21 = 0;
    (x-7)(x+3) = 0, hence x = 7 or -3

    Good luck
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  3. #3
    Member pflo's Avatar
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    Quote Originally Posted by Mhockey02 View Post
    log4^(x+7)-log4^(x-3)

    I may have figured out this one...

    (x+7)-(x-3)
    x=-(-7-3)
    x=10
    Is this right???,
    You must have an equation to end up with an equation. Since you don't know what the original expression equals, so you can't solve it. The expression, however, can be simplified.

    But all these probelems are asking about properties of logs. With logs, there are three properties you should know (in addition to the change base formula). These are:
    log_b(a^x) = x log_b(a)
    log_b(m) + log_b(n) = log_b(mn)
    log_b(m) - log_b(n) = log_b(\frac{m}{n})

    Your expression,
    log[4^{x+7}]-log[4^{x-3}] can become log[\frac{4^{x+7}}{4^{x-3}}]. Of course, when you are dividing with a common base, you can subtract exponents. Doing that would give you log[4^{(x+7)-(x-3)} or log[4^{10}].

    Plugging this into a calculator, I get log[4^{x+7}]-log[4^{x-3}]=6.0205999 (approximately).

    Quote Originally Posted by Mhockey02 View Post
    logc^x+logc^y
    You can use similar logic for this one. Another way to do both of these problems is to apply the first property (the 'exponents' property) first:

    log[c^x]+log[c^y]=xlog(c)+ylog(c)=(x+y) log(c)

    Quote Originally Posted by Mhockey02 View Post
    also this one, but it's different from the first two i think...any help getting started would be terrific!

    log2^(x+1)+log2^(x-5)=4

    I may have figured out the last problem...

    log2^(x+1)+log2^(x-5)=4
    log2x+log2(1)=log2+x+log2-5=4
    x+1+x-5=4
    x-4=4
    x=0
    ??? Is this right ???
    Nope, not correct.

    log[2^{x+1}]+log[2^{x-5}]=4 (given)
    (x+1)log(2)+(x-5)log(2)=4 ('exponents' property)
    x+1+x-5=\frac{4}{log(2)} (dividing by log(2))
    2x-4=\frac{4}{log(2)} (simplifying)
    2x=\frac{4}{log(2)}+4 (adding 4)
    x=\frac{2}{log(2)}+2 (dividing by 2)
    This is approximately x=8.644
    Last edited by pflo; March 23rd 2010 at 11:06 AM. Reason: fix latex error
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  4. #4
    Member pflo's Avatar
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    Quote Originally Posted by spacemonkey View Post
    Log2(x+1) + log2(x-5) = 4;
    Log2((x+1)(x-5)) = 4;
    Log2(x^2 - 4x - 5) = 4;
    x^2 - 4x - 5 = 2^4;
    x^2 - 4x - 5 = 16;
    x^2 - 4x - 21 = 0;
    (x-7)(x+3) = 0, hence x = 7 or -3
    If this IS what is meant by the problem (log base 2), then this is a good method. Be careful, however, because you CAN get extraneous solutions. In this case, the -3 is extraneous. You cannot have a log of 0 or the log of a negative. If x=-3, then you've got the log of -2 and the log of -8 in the original equation. This doesn't work. The only solution that works is 7.
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  5. #5
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    Quote Originally Posted by pflo View Post
    You cannot have a log of 0 or the log of a negative. If x=-3, then you've got the log of -2 and the log of -8 in the original equation. This doesn't work. The only solution that works is 7.
    yes you are absolutely right
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