Help Please. 4^2x+3=1

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Help Please. 4^2x+3=1

Algebra
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I need help badly in couple of questions, someone please help!!
is the problem \[\Large 4^{2x}+3=1\] OR is it \[\Large 4^{2x+3}=1\]
yeah that second one

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Hint: \[\Large 4^{2x+3}=1\] \[\Large 4^{2x+3}=4^0\]
so we are not using that 1?
we are, just a different form of it 4^0 is the same as 1
ok
the bases are the same (4) so the exponents must be the same
so 2x + 3 = 0
yup i Got the answer thanks :)
ok great
can you please help me with couple of them too?
sure a few more
ok here is another one
\[10^{-3x}\times10^{x}=\frac{ 1 }{ 10 }\]
\[10^{-3x}\times10^{x}=\frac{ 1 }{ 10 }\] \[10^{-3x}\times10^{x}=10^{-1}\] \[10^{-3x+x}=10^{-1}\] I'll let you finish
ok
what do you get
\[x=\frac{ 1 }{ 2 }\]
I need help with couple more pleaseeee!!
\[64\times16^{-3x}=16^{3x-2}\]
-3x + x = -1 -2x = -1 x = 1/2 good
thanks
\[64\times16^{-3x}=16^{3x-2}\] \[(4^3)\times(4^2)^{-3x}=(4^2)^{3x-2}\] \[(4^3)\times4^{2(-3x)}=(4)^{2(3x-2)}\] \[(4^3)\times4^{-6x}=(4)^{6x-4}\] \[4^{3+(-6x)}=(4)^{6x-4}\] \[4^{3-6x}=(4)^{6x-4}\]
I got an answer: \[x=\frac{ 7 }{ 12 }\]
good
\[\frac{ 81^{3n+2} }{ 243^{-n} }=3^{4}\]
oh maybe I can do this one let me try first :)
alright, tell me what you get
I got \[-\frac{ 4 }{ 17 }\]
you nailed it
:)
\[81\times9^{-2b-2}=27\]
tell me what you get
ok
oh I got that one \[-\frac{ 3 }{ 4 }\]
good, that's correct
ok
\[\left(\begin{matrix}1 \\ 6\end{matrix}\right)^{3x+2}\times216^{3x}=\frac{ 1 }{ 216 }\]
and what did you get
I m having trouble starting it but still trying
\[\Large \left(\frac{1}{6}\right)^{3x+2}\times216^{3x}=\frac{ 1 }{ 216 }\] \[\Large \left(6^{-1}\right)^{3x+2}\times(6^3)^{3x}=6^{-3}\] \[\Large \left(6\right)^{-(3x+2)}\times(6)^{9x}=6^{-3}\] \[\Large 6^{-(3x+2)+9x}=6^{-3}\]
oh I only messed up with one sign

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