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Why does Wolfram Alpha say that $\underbrace{\ln(y(y^{2}-4))}_\text{EQ1} \neq \underbrace{\ln(y) + \ln(y-2) + \ln(y+2)}_\text{EQ2}$ (seen here)?

The Produce Rule of Logarithms states that: $$\log_b{(M*N)}=\log_b{M} + \log_b{N}$$

Therefore, $$ \begin{align} \ln(y) + \ln(y-2) + \ln(y+2)& = \\ \ln(y) + \ln((y-2)(y+2)) & = \\ \ln(y) + \ln(y^2-4) & = \ln(y(y^2 - 4)) = \mathbf{EQ1} \ \checkmark \end{align} $$

However, Wolfram Alpha is saying that EQ1 "is not always equal to" EQ2. The graph Wolfram provides feels problematic, because it is showing two dramatically different curves when in reality both sizes are the same curve. What am I missing here?

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  • $\begingroup$ Likely it’s assuming that $y$ is complex. $\endgroup$ – amd Aug 29 '18 at 9:32
  • $\begingroup$ What do you expected e.g. for $y=-3?$ $\endgroup$ – gammatester Aug 29 '18 at 9:34
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    $\begingroup$ For $y = -1$, the LHS is well defined, but for the RHS, none of them are valid when considering $y \in \mathbb R$. $\endgroup$ – xbh Aug 29 '18 at 9:36
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    $\begingroup$ Or when $y=-1$ ? (oops - xbh just said that !) $\endgroup$ – gandalf61 Aug 29 '18 at 9:36
  • $\begingroup$ In Mathematica, "Simplify[Log[y (y^2 - 4)] - Log[y] - Log[y - 2] - Log[y + 2], Assumptions -> y > 2]" works, but I don't know how to do this in Alpha. $\endgroup$ – awkward Aug 29 '18 at 12:23
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For example: $\ln(y(y^2-4))$ is defined for $y=-1$, but $\ln(y) + \ln(y-2) + \ln(y+2)$ is not defined for $y=-1$.

The rule $\log_b{(M*N)}=\log_b{M} + \log_b{N}$ is only valid if both $M$ and $N$ are positive !

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  • $\begingroup$ Quick question, if I was to encapsulate each interior with absolute values (i.e. $\ln|y(y^{2}-4)| \stackrel{?}{=} \ln|y| + \ln|y-2| + \ln|y+2|$), why is it still not always equal to each other? $\endgroup$ – Danny Dan Aug 29 '18 at 10:17

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