I cannot understand the result from logarithms manipulations even though I am going over logarithmic properties. I am simply stuck. So here is the problem: $$n = 2^k \implies k = \log_2n$$ $$x = 3^k \implies k = \log_3x$$ $$\log_2n = \frac{\log_3n}{\log_32}$$ $$\log_3x = \log_3n^{1.59}$$

I do not understand how we got the last line of the solution! I understand it must be very simple to understand, but I am stuck with it. I will be thankful for help.

  • 1
    $\begingroup$ I bet that $\log_23\approx1.59$ $\endgroup$ – barak manos Jan 25 '15 at 18:42

$$\log_3x = \frac{\log_3 n}{\log_3 2} = 1.59 \log_3 n$$

Because $\frac{1}{\log_3 2} = 1.59$ and use that $c\log_a b = \log_ab^{c}$.


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