anonymous
  • anonymous
if 289=17^1/5x then x=?
Mathematics
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SOLVED
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chestercat
  • chestercat
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Yttrium
  • Yttrium
is the equation like this \[289=17^{1/5x}\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
Hey ! What is the question???????????

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anonymous
  • anonymous
289=17power 1/5x
anonymous
  • anonymous
Oh ! OK !
Yttrium
  • Yttrium
\[289 = 17^{1/5x}\] \[289 = 17^{2}\] \[17^{2} = 17^{1/5x}\] What you need to do is to equate the exponents since they have the same base of 17 Can you do it now?
anonymous
  • anonymous
@:Yttrium : You re falllllllllllllllllllllllllllllllllllllllllllll
anonymous
  • anonymous
s0 10 is the answer
anonymous
  • anonymous
NO :)
Yttrium
  • Yttrium
1/10, I guess
anonymous
  • anonymous
NO :)
Yttrium
  • Yttrium
10x = 1, Therefore x = 1/10
Yttrium
  • Yttrium
@jimra Did you get it?
anonymous
  • anonymous
i also get the same answer .but is it right?
Yttrium
  • Yttrium
Yes, you can check that by substituting the value of x. Check whether the equation will yield to 289 = 289
anonymous
  • anonymous
right
anonymous
  • anonymous
HEY look ! First answer this : 17^x=289 What is the x ?!
Yttrium
  • Yttrium
2
zzr0ck3r
  • zzr0ck3r
ahh the x is in the denominator...
Yttrium
  • Yttrium
Yes.
anonymous
  • anonymous
Ok ! Now 2/1.5 =? ?=x !!!
anonymous
  • anonymous
x in numerator
Yttrium
  • Yttrium
if x in numerator, therefore 2 = x/5 Hence x = 10 Maybe, i just misinterpret the eq'n. I'm sorry
zzr0ck3r
  • zzr0ck3r
\[17^2=17^{\frac{x}{5}}\]????
anonymous
  • anonymous
Oh ! OK ! SAY : Your answer is right ! 2.5=10
zzr0ck3r
  • zzr0ck3r
\[17^2=17^{\frac{1}{5}x}\\log_{17}(17^2)=log_{17}{(17^{\frac{1}{5}x}})\\2=\frac{1}{5}x\\x=10\]
anonymous
  • anonymous
@ Yuttrium : You are true ! @ jimra : The answer is 10 . !!!
zzr0ck3r
  • zzr0ck3r
i thought the answer was 2.5 = 10
Yttrium
  • Yttrium
@jimra, Did you get the concept?
anonymous
  • anonymous
@jimra , Answer to Yttrium question !
goformit100
  • goformit100
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