JJenk
  • JJenk
Solve. open parentheses square root of 7 close parentheses to the 6 x power = 49x-6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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marissalovescats
  • marissalovescats
Can you draw this please?
marissalovescats
  • marissalovescats
|dw:1373139028363:dw|
JJenk
  • JJenk
|dw:1373139009699:dw|

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More answers

marissalovescats
  • marissalovescats
Hm... I recognize this. Let me go look at past notes.
JJenk
  • JJenk
These were the choices i was given x = negative 21 over 2 x = -6 x = negative 6 over 5 x = -12
marissalovescats
  • marissalovescats
Does this happen to be from a lesson called "Solve Exponential Equations" ?
JJenk
  • JJenk
Yes
marissalovescats
  • marissalovescats
Okay.. Uhm.. I've never seen one with a square root but let me see if I can solve it then I'll show you
Jhannybean
  • Jhannybean
\[\large \sqrt{(7)}^{6x}=49^{x-6}\]\[\large (\sqrt{7})^{6x} = \color{green}{(\sqrt{7})^{4(x-6)}}\]\[\large 6x = 4(x-6)\]\[\large 6x=4x-24\] solve for x.
marissalovescats
  • marissalovescats
No^^^ Its not 7^4. It's 7^2.
marissalovescats
  • marissalovescats
I got -6/5 and I'll show you.
marissalovescats
  • marissalovescats
|dw:1373139792425:dw|
Jhannybean
  • Jhannybean
i changed it to the sae base.\(\sqrt{7}^4\)=49. and you've gotthe base \(\sqrt{7}\) on the left side.
Jhannybean
  • Jhannybean
Tell me how you changed \(\large (\sqrt{7})^{6x}\) to just 7?
marissalovescats
  • marissalovescats
I just figured because you square it. I've never done one like this. I was going out on a limb. But you are are correct because qroot7^4 is 49.
marissalovescats
  • marissalovescats
So in that case it's -12 and not -6/5.
primeralph
  • primeralph
@marissalovescats @Jhannybean Same thing.
primeralph
  • primeralph
|dw:1373140157416:dw|
marissalovescats
  • marissalovescats
Yes but the way she solved it was the correct way.
primeralph
  • primeralph
|dw:1373140211718:dw|
primeralph
  • primeralph
|dw:1373140263170:dw|
marissalovescats
  • marissalovescats
I just put 7^2 because I knew you have to have like the same "base number" Or whatever.
primeralph
  • primeralph
That 's if you want to use 7 and not sqrt(7)
Jhannybean
  • Jhannybean
\[\large (\sqrt{7})^4 = 7^{4/2} = 7^2 \]
Jhannybean
  • Jhannybean
Yeah....same thing.

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