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Math Help - Please help with difference of logarithms

  1. #1
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    Question Please help with difference of logarithms

    Please help me with this problem. I've tried at least 6 different methods to solve it and they've ALL been wrong. I'm wasting too much time on this problem.

    Here it is:

    log5(x+1) - log4(x-2) = 1

    Where log5 and log4 are log base 5 and log base 4, respectively. I have tried converting to base 4, tried converting to base 5, and neither has worked. If you can, please take me through it step by step.

    According to my book, the answer is supposed to be x = 2.79, and I haven't gotten close to it yet.

    Please help me with this problem, I've spent HOURS on it and I'm normally very good with logs and exponents...I REALLY want to get this right in case I see it on an exam. this can't be that complicated so what the hell am I doing wrong? I can't believe I'm having this much trouble with such a simple problem
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  2. #2
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    \displaystyle\ log_5(x+1)=\frac{log_4(x+1)}{log_45}=\frac{log_4(x  +1)}{\left(\frac{ln5}{ln4}\right)}

    =\displaystyle\frac{log_4(x+1)}{1.16}

    Then...

    log_4(x+1)-1.16log_4(x-2)=1.16

    log_4(x+1)=1.16[1+log_4(x-2)]=1.16[log_44+log_4(x-2)]=1.16[log_4(4x-8)]

    Solving for this graphically gives x=2.788
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  3. #3
    A Plied Mathematician
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    I'm afraid it looks as though an analytical solution does not exist. Here's a numerical solution.
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  4. #4
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    Thank you for the quick replies.

    It turns out the problem is meant to be solved graphically, according to the textbook...only thing is that my professor ALWAYS tells us to ignore the directions and solve everything by hand. I checked with other sources and they also said I can't do this by hand, so it looks like the professor's mistake... the guy is nice but a huge flake. I got as far as Archie did and then couldn't get any further. Now I know why. Thanks again.
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  5. #5
    A Plied Mathematician
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    You're welcome for my contributions. Have a good one!
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