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

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

Mathematics
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Can you draw this please?
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Other answers:

Hm... I recognize this. Let me go look at past notes.
These were the choices i was given x = negative 21 over 2 x = -6 x = negative 6 over 5 x = -12
Does this happen to be from a lesson called "Solve Exponential Equations" ?
Yes
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
\[\large \sqrt{(7)}^{6x}=49^{x-6}\]\[\large (\sqrt{7})^{6x} = \color{green}{(\sqrt{7})^{4(x-6)}}\]\[\large 6x = 4(x-6)\]\[\large 6x=4x-24\] solve for x.
No^^^ Its not 7^4. It's 7^2.
I got -6/5 and I'll show you.
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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.
Tell me how you changed \(\large (\sqrt{7})^{6x}\) to just 7?
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.
So in that case it's -12 and not -6/5.
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Yes but the way she solved it was the correct way.
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I just put 7^2 because I knew you have to have like the same "base number" Or whatever.
That 's if you want to use 7 and not sqrt(7)
\[\large (\sqrt{7})^4 = 7^{4/2} = 7^2 \]
Yeah....same thing.

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