If $frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15;$ then which of the following equals...












2












$begingroup$


The problem is:




If $frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15,;$ then which of the following equals $a×b×c$ ?



A) $frac {1}{375}$



B) $frac{1}{125}$



C) $frac{27}{125}$



D) $frac{3}{125}$



E) $frac{27}{5}$




I think the question is wrong.



My counterexample:




Let $x=m,; y=m-3,; z=m.$



Then $a=b=c=frac{3^{m-1}}{5}.$



So, $a×b×c=frac{3^{3m-3}}{125},; minmathbb{R}.$




Am I right?






algebra-precalculus proof-verification contest-math examples-counterexamples exponentiation







share|cite|improve this question











$endgroup$












  • $begingroup$
    "$a times b times c$ which of the following" maybe you need a verb for this sentence? :P
    $endgroup$
    – L. F.
    Dec 14 '18 at 8:28










  • $begingroup$
    @L.F. Or maybe only an "is" ;P
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 11:54








  • 3




    $begingroup$
    @ary Technically "is" is also a verb XD
    $endgroup$
    – L. F.
    Dec 14 '18 at 11:55
















2












$begingroup$


The problem is:




If $frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15,;$ then which of the following equals $a×b×c$ ?



A) $frac {1}{375}$



B) $frac{1}{125}$



C) $frac{27}{125}$



D) $frac{3}{125}$



E) $frac{27}{5}$




I think the question is wrong.



My counterexample:




Let $x=m,; y=m-3,; z=m.$



Then $a=b=c=frac{3^{m-1}}{5}.$



So, $a×b×c=frac{3^{3m-3}}{125},; minmathbb{R}.$




Am I right?






algebra-precalculus proof-verification contest-math examples-counterexamples exponentiation







share|cite|improve this question











$endgroup$












  • $begingroup$
    "$a times b times c$ which of the following" maybe you need a verb for this sentence? :P
    $endgroup$
    – L. F.
    Dec 14 '18 at 8:28










  • $begingroup$
    @L.F. Or maybe only an "is" ;P
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 11:54








  • 3




    $begingroup$
    @ary Technically "is" is also a verb XD
    $endgroup$
    – L. F.
    Dec 14 '18 at 11:55














2












2








2





$begingroup$


The problem is:




If $frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15,;$ then which of the following equals $a×b×c$ ?



A) $frac {1}{375}$



B) $frac{1}{125}$



C) $frac{27}{125}$



D) $frac{3}{125}$



E) $frac{27}{5}$




I think the question is wrong.



My counterexample:




Let $x=m,; y=m-3,; z=m.$



Then $a=b=c=frac{3^{m-1}}{5}.$



So, $a×b×c=frac{3^{3m-3}}{125},; minmathbb{R}.$




Am I right?






algebra-precalculus proof-verification contest-math examples-counterexamples exponentiation







share|cite|improve this question











$endgroup$




The problem is:




If $frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15,;$ then which of the following equals $a×b×c$ ?



A) $frac {1}{375}$



B) $frac{1}{125}$



C) $frac{27}{125}$



D) $frac{3}{125}$



E) $frac{27}{5}$




I think the question is wrong.



My counterexample:




Let $x=m,; y=m-3,; z=m.$



Then $a=b=c=frac{3^{m-1}}{5}.$



So, $a×b×c=frac{3^{3m-3}}{125},; minmathbb{R}.$




Am I right?






algebra-precalculus proof-verification contest-math examples-counterexamples exponentiation




algebra-precalculus proof-verification contest-math examples-counterexamples exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 16:10









user376343

3,9234829




3,9234829










asked Dec 14 '18 at 8:04









ElementaryElementary

361111




361111












  • $begingroup$
    "$a times b times c$ which of the following" maybe you need a verb for this sentence? :P
    $endgroup$
    – L. F.
    Dec 14 '18 at 8:28










  • $begingroup$
    @L.F. Or maybe only an "is" ;P
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 11:54








  • 3




    $begingroup$
    @ary Technically "is" is also a verb XD
    $endgroup$
    – L. F.
    Dec 14 '18 at 11:55


















  • $begingroup$
    "$a times b times c$ which of the following" maybe you need a verb for this sentence? :P
    $endgroup$
    – L. F.
    Dec 14 '18 at 8:28










  • $begingroup$
    @L.F. Or maybe only an "is" ;P
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 11:54








  • 3




    $begingroup$
    @ary Technically "is" is also a verb XD
    $endgroup$
    – L. F.
    Dec 14 '18 at 11:55
















$begingroup$
"$a times b times c$ which of the following" maybe you need a verb for this sentence? :P
$endgroup$
– L. F.
Dec 14 '18 at 8:28




$begingroup$
"$a times b times c$ which of the following" maybe you need a verb for this sentence? :P
$endgroup$
– L. F.
Dec 14 '18 at 8:28












$begingroup$
@L.F. Or maybe only an "is" ;P
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 11:54






$begingroup$
@L.F. Or maybe only an "is" ;P
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 11:54






3




3




$begingroup$
@ary Technically "is" is also a verb XD
$endgroup$
– L. F.
Dec 14 '18 at 11:55




$begingroup$
@ary Technically "is" is also a verb XD
$endgroup$
– L. F.
Dec 14 '18 at 11:55










1 Answer
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$begingroup$

Yes some information is missing, indeed we have that



$$frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15$$



then



$$abc=frac{3^{(x+y+z)}}{125}$$



then we need a condition for $t=x+y+zin mathbb R$, since $3^{t}$ can assume any value $in(0,infty)$.






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    active

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    $begingroup$

    Yes some information is missing, indeed we have that



    $$frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15$$



    then



    $$abc=frac{3^{(x+y+z)}}{125}$$



    then we need a condition for $t=x+y+zin mathbb R$, since $3^{t}$ can assume any value $in(0,infty)$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Yes some information is missing, indeed we have that



      $$frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15$$



      then



      $$abc=frac{3^{(x+y+z)}}{125}$$



      then we need a condition for $t=x+y+zin mathbb R$, since $3^{t}$ can assume any value $in(0,infty)$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Yes some information is missing, indeed we have that



        $$frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15$$



        then



        $$abc=frac{3^{(x+y+z)}}{125}$$



        then we need a condition for $t=x+y+zin mathbb R$, since $3^{t}$ can assume any value $in(0,infty)$.






        share|cite|improve this answer











        $endgroup$



        Yes some information is missing, indeed we have that



        $$frac {a}{3^{x-1}}=frac{b}{3^{y+2}}=frac{c}{3^{z-1}}=frac 15$$



        then



        $$abc=frac{3^{(x+y+z)}}{125}$$



        then we need a condition for $t=x+y+zin mathbb R$, since $3^{t}$ can assume any value $in(0,infty)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 8:24

























        answered Dec 14 '18 at 8:18









        gimusigimusi

        92.9k84494




        92.9k84494






























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