JJenk
Solve. open parentheses square root of 7 close parentheses to the 6 x power = 49x6



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marissalovescats
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Can you draw this please?

marissalovescats
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dw:1373139028363:dw

JJenk
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dw:1373139009699:dw

marissalovescats
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Hm... I recognize this. Let me go look at past notes.

JJenk
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These were the choices i was given
x = negative 21 over 2
x = 6
x = negative 6 over 5
x = 12

marissalovescats
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Does this happen to be from a lesson called "Solve Exponential Equations" ?

JJenk
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Yes

marissalovescats
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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
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\[\large \sqrt{(7)}^{6x}=49^{x6}\]\[\large (\sqrt{7})^{6x} = \color{green}{(\sqrt{7})^{4(x6)}}\]\[\large 6x = 4(x6)\]\[\large 6x=4x24\] solve for x.

marissalovescats
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No^^^ Its not 7^4. It's 7^2.

marissalovescats
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I got 6/5 and I'll show you.

marissalovescats
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dw:1373139792425:dw

Jhannybean
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i changed it to the sae base.\(\sqrt{7}^4\)=49. and you've gotthe base \(\sqrt{7}\) on the left side.

Jhannybean
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Tell me how you changed \(\large (\sqrt{7})^{6x}\) to just 7?

marissalovescats
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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
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So in that case it's 12 and not 6/5.

primeralph
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@marissalovescats @Jhannybean
Same thing.

primeralph
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dw:1373140157416:dw

marissalovescats
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Yes but the way she solved it was the correct way.

primeralph
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dw:1373140211718:dw

primeralph
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dw:1373140263170:dw

marissalovescats
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I just put 7^2 because I knew you have to have like the same "base number" Or whatever.

primeralph
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That 's if you want to use 7 and not sqrt(7)

Jhannybean
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\[\large (\sqrt{7})^4 = 7^{4/2} = 7^2 \]

Jhannybean
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Yeah....same thing.