log_{3}(x) + log_{9}(x) = 12 solve for x​

Question

log_{3}(x) + log_{9}(x) = 12
solve for x​

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Bình An 5 years 2021-09-03T10:21:51+00:00 1 Answers 19 views 0

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  1. Eirian
    0
    2021-09-03T10:23:46+00:00
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    Answer:

      x = 3⁸  

    Step-by-step explanation:

    Step(i):-

    Given that

                  log _{3} (x)+ log_{9} (x) =12

                 log _{3} (x)+ log_{3^{2} } (x) =12

      we know that

            log^{a} _{b} = \frac{loga}{logb}

             log _{3} (x)+ \frac{logx}{log3^{2} } =12

    Step(ii):-

    Apply log xⁿ = nlogx

          log _{3} (x)+\frac{1}{2} \frac{logx}{log3 } =12

        log _{3} (x)+ \frac{1}{2} log_{3 } (x) =12

       log _{3} (x)+ log_{3 } (x)^{\frac{1}{2} } =12      ( ∵  log xⁿ = nlogx)

     Apply log(ab) = loga+logb

     log _{3} (x (x^{\frac{1}{2} }) =12

      log _{3} ( (x^{\frac{3}{2} }) =12

    \frac{3}{2} log _{3} ( x) =12

     \frac{1}{2} log _{3} ( x) = 4

    log _{3} ( x) = 8

    we know that  log _{b} ( x) = a    ⇒  x = bᵃ

    ∴    x = 3⁸  

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