Let g(x) = log5|2log3|. Find the product of the zeros of g.
I did not understand how this function could have zeros as there is no unknown to calculate for. I'm also not sure what the purpose of the absolute value is for.
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$\log_5\lvert2\log_3\rvert$ has no sense since $\log_3$ has no argument. – manooooh Dec 08 '19 at 05:07
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In think there is a misprint. The definition should be $g(x)=log_5|2log_3 x|$. This is $0$ iff $2log_3 x=\pm 1$. This means $x=3^{1/2} $ or $x =3^{-1/2}$ and the product of these zeros is $1$.
Kavi Rama Murthy
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Lol a good answer before the very last step... P.S. even without calculation we know that $x$ is a zero iff $x^{-1}$ is a zero, so the product (if makes sense) must be $1$. – WhatsUp Dec 08 '19 at 05:41
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How does dividing 0 by 5 give ±1? (is there any particular rule of logs applied here) – V11 Dec 08 '19 at 06:51
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@V11 $\log_a x=0$ iff $x=1$ for any base $a$. Hence $\log_5|2log_3 x|=0$ iff $|2log_3 x|=1$ which means $2\log_3 x =\pm 1$. – Kavi Rama Murthy Dec 08 '19 at 11:26
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