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Math Help - log/exponent problem

  1. #1
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    Question log/exponent problem

    The question consists of one line only, but the silly me failed to solve it.
    Please find the value of x for this equation:
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  2. #2
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    I cannot say for anyone else, but I have absolutely no idea what that says.
    For example, what could ^4\log 2 mean?
    Let alone all the rest of the jumble of notation.

    Why not learn to post in symbols? You can use LaTeX tags
    [tex] \log_{4}(2) [/tex] gives \log_{4}(2)
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  3. #3
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    aww, I am terribly sorry.
    In my stupid country, \log_{n} is written as ^{n}log instead.

    So, it should be written like this for international standard:

    \frac{6}{5}(h^{\log_{h}x.\log_{10}h.\log_{h}5})-3^{\log\frac{x}{10}}=9^{\log_{100}x+\log_{4}2}
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  4. #4
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    Quote Originally Posted by eternia View Post
    aww, I am terribly sorry.
    In my stupid country, \log_{n} is written as ^{n}log instead.So, it should be written like this for international standard:
    \frac{6}{5}(h^{\log_{h}x.\log_{10}h.\log_{h}5})-3^{\log\frac{x}{10}}=9^{\log_{100}x+\log_{4}2}
    I am sorry to tell you, that reply is useless.
    What does tho O.P. mean?
    What was included in to O.P. is gibberish.
    What do you mean?
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  5. #5
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    OK so I got the problem as:

    Find the value(s) of x:

    \displaystyle \frac{6(h^{\log_{h}(x) \, \cdot \, \log_{10}(h) \, \cdot \, \log_{h}(5)})}{5} - 3^{\log(\frac{x}{10})}=9^{\log_{100}(x)+\log_{4}(2  )}


    For the 3^{\log\frac{x}{10}} part, what is the base of this logarithm? Is it supposed to be log_e?

    To solve for x, notice how the h's can cancel out. The first step would be to cancel out the h's so you don't have 2 variables.
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  6. #6
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    Question

    The one above me figured out my equation properly,
    but for that one, the base number is 10, not e
    If it's e, it would be written as Ln instead, no?
    So, it's :

    3^{\log_{10}(\frac {x} {10})

    For starting, we cancel out the h of course, but I am still having difficulties....
    You see, even though the 9 can be turned into 3^2
    But I can't find a way to trash the 5, which is a prime number on it's own.
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  7. #7
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    I don't think you can solve this algebraically.

    Maybe you could use an approximation method to solve for x instead?
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  8. #8
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    My prof. said that it can be solved, x = 100
    Our task is to show him the process....
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  9. #9
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    This is quite tricky.
    Split this equation up into 3 pieces:

    \displaystyle \underbrace{\tfrac{6}{5}(h^{\log_{h}(x) \, \cdot \, \log_{10}(h) \, \cdot \, \log_{h}(5)})}_{\text{Part a}} - \underbrace{3^{\log_{10}(\frac{x}{10})}}_\text{Par  t b}}=\underbrace{9^{\log_{100}(x)+\log_{4}(2)}}_{\t  ext{Part c}}

    I'll skip out some parts or else it will be too long.
    Part a can be expressed as:

    \displaystyle \tfrac{6}{5}(h^{\log_{h}(x) \, \cdot \, \log_{10}(h) \, \cdot \, \log_{h}(5)})} = 6 \times 5^{\log_{10}(x) - 1}

    Part b can be expressed as:

    3^{\log_{10}(\frac{x}{10})} = 3^{\log_{10}(x) - 1}

    Part c can be expressed as:

    9^{\log_{100}(x)+\log_{4}(2)} = 3^{\log_{10}(x) + 1}

    Hence the equation becomes:

    \displaystyle 6 \times 5^{\log_{10}(x) - 1} -3^{\log_{10}(x) - 1} = 3^{\log_{10}(x) + 1}

    \displaystyle 6 \times 5^{\log_{10}(x) - 1} = \underbrace{3^{\log_{10}(x) + 1} + 3^{\log_{10}(x) - 1}}_{\text{Simplify}}

    To simplify:

    3^{\log_{10}(x)} \times 3^1 + 3^{\log_{10}(x)} \times 3^{-1}

    3^{\log_{10}(x)} ( 3^1 + 3^{-1})

    3^{\log_{10}(x)} (\frac{10}{3}) \underbrace{\times 3 \div 3}_{\text{Multiply by 1}}

    3^{\log_{10}(x)} \div 3 \times (\frac{10}{3}) \times 3

    3^{\log_{10}(x)-1} \times 10

    Thus the equation becomes:

    \displaystyle 6 \times 5^{\log_{10}(x) - 1} =10 \times 3^{\log_{10}(x)-1}

    To make it easier, let u = \log_{10}(x)

    \displaystyle 6 \times 5^{u - 1} =10 \times 3^{u-1}

    Now use logarithms to solve for u, the thus solve for x.
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  10. #10
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    Eureka!
    I am forever indebted to you, big bro.
    Thank you, thank you so much.
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