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Math Help - I'm stuck on logarithmic equations :(

  1. #1
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    I'm stuck on logarithmic equations :(

    I had a 98 in my Algebra class until this chapter came up. It's the last math course I ever have to take so I am hoping to finish strong.

    Typically when a specific multiplier or divisor or whatever is present in every part of an equation, it can simply be removed, right? So if I have this equation:

    log24x + log29 = log218

    Am I allowed to simply remove the log portion from the equation and leave it was 4x+9=18? It just doesn't seem right. If not how do I solve it?

    Is it the same with natural logs, since they always have the same base? Take

    lnx- ln(x-4) = ln3

    Thanks in advance.
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  2. #2
    Senior Member Twig's Avatar
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    hi

    Hi!

    I would suggest that you take a good look at the logharitmic laws, and exponent laws as well.
    You simply must now then well enough in order to solve these types of equations.

    log2 (4x) = log 2 (4) + log2 (x)

    These are one of the laws.

    Another example: lg (10/3) = lg (10) - lg(3) = 1 - lg(3)

    This is the second law.

    I really really think you´ll easy crack these once you´ve mastered this laws.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by gregorywb View Post
    I had a 98 in my Algebra class until this chapter came up. It's the last math course I ever have to take so I am hoping to finish strong.

    Typically when a specific multiplier or divisor or whatever is present in every part of an equation, it can simply be removed, right? So if I have this equation:

    log24x + log29 = log218

    Am I allowed to simply remove the log portion from the equation and leave it was 4x+9=18? It just doesn't seem right. If not how do I solve it?


    Since the logarithms have the same base, we can combine them: \log_a(b)+\log_a(c)=\log_a(bc)

    Thus, \log_2(4x)+\log_2(9)=\log_2(18)\implies \log_2(36x)=\log_2(18).

    Now we can let the logs drop out of the equation to get 36x=18\implies x=...


    Quote Originally Posted by gregorywb View Post
    Is it the same with natural logs, since they always have the same base? Take

    lnx- ln(x-4) = ln3

    Thanks in advance.
    Yes. Since you are subtracting natural logarithms: \ln(a)-\ln(b)=\ln\bigg(\frac{a}{b}\bigg)

    Thus, \ln(x)-\ln(x-4)=\ln(3)\implies \ln\bigg(\frac{x}{x-4}\bigg)=\ln(3).

    Now the ln(s) can drop out and we are left with \frac{x}{x-4}=3

    Can you take it from here?

    I hope that this makes more sense now!

    --Chris
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