How do I square $\log_2(3)$. Does it become $2\log_2(3)$ ?
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asked May 7, 2015 at 10:19
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7 Answers
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No. $(\log_2(3))^2$ can't be simplified.
However, $\log_2(3^2)=2\log_2(3)$.
answered May 7, 2015 at 10:21
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38$\begingroup$ Well, one does have $(\log_2(3))^2=\log_2(3^{\log_2(3)})$, but I'm not sure whether you can call that a simplification. $\endgroup$ Commented May 7, 2015 at 14:09
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No, it doesn't. Logarithms follow this rule: $$ \log_b (a^c) = c\log_b a, $$ while your statement says that $$ (\log_b a)^c=c\log_b a,$$ which is basically saying the same as $x^y=yx$.
answered May 7, 2015 at 10:24
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$\log_2(3) \approx 1.58496$ as you can easily verify.
$(\log_2(3))^2 \approx (1.58496)^2 \approx 2.51211$.
$2 \log_2(3) \approx 2 \cdot 1.58496 \approx 3.16992$.
$2^{\log_2(3)} = 3$.
Do any of those appear to be equal?
(Whenever you are wondering whether some general algebraic relationship holds, it's a good idea to first try some simple numerical examples to see if it is even possible.)
Actually, the only way that $(\log_2(3))^2 = 2 \log_2(3)$ could hold is if $\log_2(3)$ were equal to 2 or 0. That is clearly false since $2^2 = 4 \ne 3$ and $2^0 = 1 \ne 3$.
answered May 7, 2015 at 21:28
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1$\begingroup$ I am an engineer, and I would say that 3rd and 4rt are equal, indeed. :P $\endgroup$ Commented May 8, 2015 at 11:56
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$\begingroup$ You: Actually, the only way that $(\log_2(3))^2 = 2 \log_2(3)$ could hold is if $\log_2(3)$ were equal to 2. Maybe that should be "2 or 0" (the solutions to $x^2=2x$). For example, $(\log_2(1))^2 = 2 \log_2(1)$. $\endgroup$ Commented May 8, 2015 at 14:48
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$\begingroup$ @JeppeStigNielsen: Thanks, fixed, $\endgroup$ Commented May 8, 2015 at 14:48
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No.The best you can do is $$(\log_2 3)^2=\log_2 3\cdot \log_2 3=\log_2 (3^{\log_2 3})$$
answered May 7, 2015 at 15:12
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$\begingroup$ log3 on base 2 cant be 2^ log 3 on base 2? $\endgroup$ Commented May 7, 2015 at 16:46
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Agreeing with the rest of the answers here, you cannot simplify any further. You could, however, do a change of base with the logs and put them in base $10$. We have the formula $$\log_bx=\frac{\log_ax}{\log_ab}$$where $a$ can be any base you want. Most common base is $10$. So we have, $$(\log_23)^2=\left(\frac{\log_{10}3}{\log_{10}2}\right)^2=\left(\frac{\log3}{\log2}\right)^2=\frac{\log^23}{\log^22}$$
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$\begingroup$ @MichaelGrünewald To make a calculator do it? $\endgroup$– KSmartsCommented May 8, 2015 at 14:50
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3$\begingroup$ @KSmarts In which world does this relate to the question asked? $\endgroup$ Commented May 8, 2015 at 14:57
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\begin{eqnarray} (\log_{2}(3))^{2} &=& (\log_{2}(3))(\log_{2}(3)) \\ &=& \log_{2}(3^{\log_{2}(3)}) \end{eqnarray} That's as simple as it gets!
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I want to add that $$ (\log_{2}(3))^{2} = 2^{\log_2 (\log_{2}(3))^{2}} = 2^{2 \log_2 (\log_{2}(3))} $$
The advantage of writing it like this is that it can be generalized to higher powers of the logarithm, e.g. $$ \ln(x)^3 = e^{3\ln(\ln(x))} $$ And for arbitrary powers of the logarithm we get $$ \ln(x)^q = \prod_{i=1}^q \ln(x) = e^{q\ln(\ln(x))} $$
