I was just wondering if i can do this.

Q. Solve $\log_{9}24=x $

$\implies9^x =24$

$\implies3^{2x}=2^3 3$

$\implies\log_3(3^{2x})= \log_3(2^3 3)$

$\implies2x=2 (3)^{1/3}$

$\implies x=3^{1/3} $

Is this actually correct or did i break some kind of log rule here? My answer appears to be out by 0.0038 compared to the book is that cause they used a calculator or is it just a fluke that my answer is so close?

  • 2
    $\begingroup$ Your method has an error in it in going from step 3 to step 4 above. On the left-hand-side you took $\log_3(3^{2x})$ to correctly equal $2x$. However, on the right-hand-side how did you conclude that $\log_3(2^33)=2(3)^{\frac{1}{3}}$? I would have just used the change of base formula to evaluate $\log_{9}24$ $\endgroup$
    – Mufasa
    Apr 19 '15 at 20:27
  • $\begingroup$ 3 twos means log base 3 will give you a two one 3 means logs base 3 will give you a cubic root of 3 but im not sure if i can do that or not. $\endgroup$
    – Faust
    Apr 19 '15 at 20:31
  • 4
    $\begingroup$ $\log_{3}2^{3}\ne2$ and $\log_{3}3\ne3^{\frac{1}{3}}$. In fact:$$\log_{3}2^{3}=3\log_{3}2$$$$\log_{3}3=1$$However, you cannot treat these two independently because:$$\log_{3}2^{3}3=\log_{3}2^3+\log_{3}3$$In general:$$\log_{b}xy=\log_{b}x+\log_{b}y$$ $\endgroup$
    – Mufasa
    Apr 19 '15 at 20:35
  • $\begingroup$ thank you for answering my question. $\endgroup$
    – Faust
    Apr 19 '15 at 20:36

Your method is incorrect as pointed out by Mufasa (sorry I didn't spot that earlier) but it's not the way I'd have done it anyway (and probably not how the book answer was obtained). This way is better:

$9^x = 24$

$\implies x\ln 9 = \ln 24$

$\implies x = \frac{\ln 24}{\ln 9} = 1.44639...$

  • $\begingroup$ perhaps that's the answer my book has, but i cant do that in my head... $\endgroup$
    – Faust
    Apr 19 '15 at 20:32
  • $\begingroup$ @Faust7 do you not have access to a calculator? Or is it a non-calculator exercise? You might be able to leave the answer as $\frac{\ln 24}{\ln 9}$ as the decimal is not exact $\endgroup$
    – imulsion
    Apr 19 '15 at 20:33

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