I've been working at this for hours. I want to know if it's even possible for the equation $n=(\frac{3}2)^k$ to be manipulated to become $n^{log_23}$.
Edit: What about $(3/2)^{log_2n}$ to become $n^{log_23}$?
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That second expression isn’t an equation so it’s not entirely clear what you mean. We have $k = log_{3/2} n$, though. – Qiaochu Yuan Oct 04 '20 at 01:05
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Thank you I appreciate the feedback! – dunlop Oct 04 '20 at 01:13
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As for your edit, in general we have $x^{\log_b y} = b^{(\log_b x)(\log_b y)} = y^{\log_b x}$ (this is probably the most bizarre and least intuitive log identity), which gives $n^{\log_2 3} = 3^{\log_2 n}$. – Qiaochu Yuan Oct 04 '20 at 01:23
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Thanks for the answer! Wouldn't we get $^{log_2(3/2)}$ instead of $^{log_23}$ since $x$ is $3/2$? – dunlop Oct 04 '20 at 03:16
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You've written down two numbers which are almost but not quite equal so there are two identities. One is $n^{\log_2 3} = 3^{\log_2 n}$ and the other is $(3/2)^{\log_2 n} = n^{\log_2 (3/2)}$. – Qiaochu Yuan Oct 04 '20 at 03:18
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Thank you so much. You've actually helped me a ton with that rule you showed me!! – dunlop Oct 04 '20 at 03:55
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You might find this 3Blue1Brown video interesting. Or the Math Exchange post that inspired it. – Paul Sinclair Oct 04 '20 at 13:51