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Math Help - Urgent help required

  1. #1
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    Step by step guiding required for logarithms sums

    Hi,
    I have encountered some challenging sums when doing my revision for my math test. Please try to help me as I will be doing the test tomorrow.

    Here are the sums:

    1)Find the exact value of x if (3x)^lg3=(4x)^lg4.

    2)Without using a calculator, evaluate (lg 5)^2 + lg 2 lg 50.

    3) Solve x in terms of a given that log base a 5 + 2= log base a (x+a) + log base a (x-3a).

    Help would be appreciated. Thanks in advance.
    Last edited by AeroScizor; February 25th 2010 at 01:26 AM.
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  2. #2
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    Hello AeroScizor

    Welcome to Math Help Forum!
    Quote Originally Posted by AeroScizor View Post
    Hi,
    I have encountered some challenging sums when doing my revision for my math test. Please try to help me as I will be doing the test tomorrow.

    Here are the sums:

    1)Find the exact value of x if (3x)^lg3=(4x)^lg4.

    2)Without using a calculator, evaluate (lg 5)^2 + lg 2 lg 50.

    3) Solve x in terms of a given that log base a 5 + 2= log base a (x+a) + log base a (x-3a).

    Help would be appreciated. Thanks in advance.
    1) Take logs of both sides:
    \log3\log(3x)=\log4\log4x

    \Rightarrow \log3(\log 3 + \log x)=\log4(\log4 + \log x)
    Solve this equation for \log x, using the 'difference of two squares' to simplify the result.

    Answer: x = \frac{1}{12}

    2) Note that \log 2 = \log 10 - \log 5 = 1 - \log 5, and that \log50 = \log10+\log5 = 1+ \log 5

    So \log2\log50 = ... (Use the difference of two squares again.)

    Answer: 1

    I'll come back to (3) later, when I have time, if no-one else has answered.

    Grandad
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  3. #3
    MHF Contributor
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    Hello again AeroScizor
    Quote Originally Posted by AeroScizor View Post
    ...
    3) Solve x in terms of a given that log base a 5 + 2= log base a (x+a) + log base a (x-3a).

    Help would be appreciated. Thanks in advance.
    Write 2 as 2\log_a(a)=\log_a(a^2), and then:
    \log_a5+\log_a(a^2) = \log_a(x+a)+\log_a(x-3a)

    \Rightarrow \log_a(5a^2)=\log_a(x+a)(x-3a)

    \Rightarrow 5a^2 = ...
    Simplify, factorise and solve the quadratic in x. Not both roots are valid though, because a,\; (x+a) and (x-3a) must all be positive.

    Answer: x = 4a

    Can you complete it now?

    Grandad
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello again AeroScizorWrite 2 as 2\log_a(a)=\log_a(a^2), and then:
    \log_a5+\log_a(a^2) = \log_a(x+a)+\log_a(x-3a)

    \Rightarrow \log_a(5a^2)=\log_a(x+a)(x-3a)

    \Rightarrow 5a^2 = ...
    Simplify, factorise and solve the quadratic in x. Not both roots are valid though, because a,\; (x+a) and (x-3a) must all be positive.

    Answer: x = 4a

    Can you complete it now?

    Grandad
    Yes, I understand now. Thanks a lot for your help! Cheers!
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