anonymous
  • anonymous
Part 1: Provide the solution to the equation 9 = sqrt3^(4x + 6). (3 points) Part 2: Using complete sentences, explain the strategy used in solving this equation the way you did. (3 points)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Shayaan_Mustafa
  • Shayaan_Mustafa
@jbobsessed You mean question is something look like this?\[9=\sqrt[3]{4x+6}\]
anonymous
  • anonymous
no the square root of 3 with an exponent of 4x+6
Shayaan_Mustafa
  • Shayaan_Mustafa
you mean?\[\sqrt{(4x+6)^{3}}\]

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More answers

anonymous
  • anonymous
no the 3 is being square rooted and the 4x+6 is the exponent..
anonymous
  • anonymous
|dw:1333128150388:dw|
Shayaan_Mustafa
  • Shayaan_Mustafa
you mean?\[9=(\sqrt{4x+6})^{3}\]
anonymous
  • anonymous
I drew it
Shayaan_Mustafa
  • Shayaan_Mustafa
oh.. ok ok.
Shayaan_Mustafa
  • Shayaan_Mustafa
so take log on bosth sides.
anonymous
  • anonymous
What is log..
Shayaan_Mustafa
  • Shayaan_Mustafa
logarithm. don't you know? if don't so how did you post this question here?
anonymous
  • anonymous
The answer is -1/2 right? if it is i just part 2 im taking a quiz.
Shayaan_Mustafa
  • Shayaan_Mustafa
ok I am solving it for you. till wait and watch me.
anonymous
  • anonymous
okay
anonymous
  • anonymous
I haven't learned log yet that's why I'm confused
Shayaan_Mustafa
  • Shayaan_Mustafa
taking log on both sides we get,\[\log9=\log((\sqrt{3})^{(4x+6)})\]or,\[\log9=(4x+6)*\log(\sqrt{3})\]since, log(9)=0.954 and log(sqrt(3))=0.238, so your equation becomes,\[0.954=(4x+6)*(0.238)\]or,\[4x+6=0.945/0.238\]or,\[4x+6=4\]or,\[4x=4-6\]or,\[4x=-2\]or,\[x=-2/4\]or, \[x=-1/2\]
anonymous
  • anonymous
Thanks! so what's part 2?
Shayaan_Mustafa
  • Shayaan_Mustafa
Part 2: Using complete sentences, explain the strategy used in solving this equation the way you did. (3 points) According to part 2, steps through which equation is solved, they must be well defined. so I did as part 2 say.
anonymous
  • anonymous
Oh okay just making sure thanks so much you're so helpful
Shayaan_Mustafa
  • Shayaan_Mustafa
you are ever welcome my dear friend.

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