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Determine all the values of real parameter $a$ so that the equation:$$(x-a)[log_4(x-5)-1]=0$$ admits a maximum number of real solutions. Thank you!

wonderingdev
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    Obviously the function $(x-a)$ can only have one root (which is $a$), regardless of what the value of $a$ is, so it remains to find the the root(s) with maximum multiplicity for $\log_4(x-5)-1$ – fixedp Apr 27 '14 at 15:23
  • This means that the root of the second equation that's 9 is respectively the value of $a$, and in such a case we'll have a maximum number of real solutions, yes? – wonderingdev Apr 27 '14 at 15:26
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    Yes, that is correct. – fixedp Apr 27 '14 at 15:29
  • But can you explain why? – wonderingdev Apr 27 '14 at 15:30
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    Say you have functions $f(x)$ and $g(x)$. Suppose $f(x)$ has $m$ roots and $g(x)$ has $n$ roots. What is the number of roots for the product $H(x)=f(x)g(x)$? – fixedp Apr 27 '14 at 15:32
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    If $A(x) \times B(x)=0$, either $A(x)=0$ or $B(x)=0$ – Claude Leibovici Apr 27 '14 at 15:33
  • @fixedp m+n roots then. – wonderingdev Apr 27 '14 at 15:38
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    Yes. Now if $f(x)=(x-a)$, we have $m=1$ regardless of what we choose for $a$. Since we want to maximise $m+n$, and $m$ is fixed, it is natural to maximise $n$. In this case, $n$ is one when $a=9$ and $n=0$ otherwise. So to maximise the number of solutions (which is $m+n$), we need to have $a=9$. – fixedp Apr 27 '14 at 15:42
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    @fixedp Thank you very much! – wonderingdev Apr 27 '14 at 15:45

1 Answers1

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$$(x-a)(\log_4(x-5)-1)=0\\(x-a)=0\lor \log_4(x-5)=1\\x=a\lor x=9$$ Now if $a\not= 9$ then solutions are $x_1=a,x_2=9$,so maximum number of solutions is for $a\not=9$

kingW3
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