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Solving logarithmic problem

  1. Sep 10, 2014 #1
    How to solve this:

    log(base16)x + log(base4)x + log(base2)x = 7

    If I have log(base16)x for example and i make

    10^(log(base16)x)

    of it, can I transform the base 16 into an exponent?

    It would look like that then:

    10^(logx)^16 = x^16

    Would that be correct?

    Then, I could solve it:

    10^(logx)^16 * 10^(logx)^4 * 10^(logx)^2 = 10^7

    that would be

    x^16 * x^4 * x^2 = 10^7

    x^22 = 10^7
    22 = log(base x)10^7

    Is that correct? Can I solve it this way?
     
  2. jcsd
  3. Sep 10, 2014 #2
    I don't (think) your method is valid, but I think using Latex in your post would help clarify your idea. Here's the normal way you tackle the problem:

    $$
    \log _{16}x+\log_{4}x+\log_{2}x=7
    $$

    If you ever calculated non-natural logs on your calculator, you know that...

    $$
    \log_{B}x = \frac{\ln x}{\ln B}
    $$

    Replace all left terms by their respective fraction:

    $$
    \frac{\ln x}{\ln 16}+\frac{\ln x}{\ln 4}+\frac{\ln x}{\ln 2}=7
    $$

    Recall that ##\ln(a^b) = b\ln a##, and that ##16## and ##4## are powers of ##2##.

    $$
    \frac{\ln x}{\ln 2^4}+\frac{\ln x}{\ln 2^2}+\frac{\ln x}{\ln 2}=7
    $$
    $$
    \frac{\ln x}{4\ln 2}+\frac{\ln x}{2\ln 2}+\frac{\ln x}{\ln 2}=7
    $$

    Find the common denominator, and consolidate the expression into one fraction:

    $$
    \frac{\ln x}{4\ln 2}+\frac{2\ln x}{4\ln 2}+\frac{4\ln x}{4\ln 2}=7
    $$
    $$
    \frac{7\ln x}{4\ln 2}=7
    $$

    Divide by ##7## to cancel them out, and multiply each side by ##4\ln 2## to isolate \ln x.

    $$
    \ln x=4\ln 2
    $$

    Recall that ##4\ln 2 = \ln 2^4 = \ln 16##:

    $$
    \ln x = \ln 16.
    $$

    Raise both sides to e to cancel out the logarithms, and you find that:

    $$
    x = 16
    $$
     
    Last edited: Sep 10, 2014
  4. Sep 10, 2014 #3

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    try converting all to log2
     
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