5
$\begingroup$

How do I square $\log_2(3)$. Does it become $2\log_2(3)$ ?

$\endgroup$
42
$\begingroup$

No. $(\log_2(3))^2$ can't be simplified.

However, $\log_2(3^2)=2\log_2(3)$.

$\endgroup$
  • 34
    $\begingroup$ 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. $\endgroup$ – Marc van Leeuwen May 7 '15 at 14:09
17
$\begingroup$

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 $x^y=yx$.

$\endgroup$
9
$\begingroup$

$\log_2(3) \approx 1.58496$ 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 $2^0 = 1 \ne 3$.

$\endgroup$
  • 1
    $\begingroup$ I am an engineer, and I would say that 3rd and 4rt are equal, indeed. :P $\endgroup$ – Ander Biguri May 8 '15 at 11:56
  • $\begingroup$ 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, $(\log_2(1))^2 = 2 \log_2(1)$. $\endgroup$ – Jeppe Stig Nielsen May 8 '15 at 14:48
  • $\begingroup$ @JeppeStigNielsen: Thanks, fixed, $\endgroup$ – Nate Eldredge May 8 '15 at 14:48
6
$\begingroup$

No.The best you can do is $$(\log_2 3)^2=\log_2 3\cdot \log_2 3=\log_2 (3^{\log_2 3})$$

$\endgroup$
  • $\begingroup$ log3 on base 2 cant be 2^ log 3 on base 2? $\endgroup$ – Raunak Banerjee May 7 '15 at 16:46
2
$\begingroup$

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, $$(\log_23)^2=\left(\frac{\log_{10}3}{\log_{10}2}\right)^2=\left(\frac{\log3}{\log2}\right)^2=\frac{\log^23}{\log^22}$$

$\endgroup$
  • 5
    $\begingroup$ For what purpose? $\endgroup$ – Michael Le Barbier Grünewald May 8 '15 at 11:26
  • $\begingroup$ @MichaelGrünewald To make a calculator do it? $\endgroup$ – KSmarts May 8 '15 at 14:50
  • 3
    $\begingroup$ @KSmarts In which world does this relate to the question asked? $\endgroup$ – Michael Le Barbier Grünewald May 8 '15 at 14:57
2
$\begingroup$

\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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.