anonymous
  • anonymous
Help Please. 4^2x+3=1
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I need help badly in couple of questions, someone please help!!
jim_thompson5910
  • jim_thompson5910
is the problem \[\Large 4^{2x}+3=1\] OR is it \[\Large 4^{2x+3}=1\]
anonymous
  • anonymous
yeah that second one

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More answers

jim_thompson5910
  • jim_thompson5910
Hint: \[\Large 4^{2x+3}=1\] \[\Large 4^{2x+3}=4^0\]
anonymous
  • anonymous
so we are not using that 1?
jim_thompson5910
  • jim_thompson5910
we are, just a different form of it 4^0 is the same as 1
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
the bases are the same (4) so the exponents must be the same
jim_thompson5910
  • jim_thompson5910
so 2x + 3 = 0
anonymous
  • anonymous
yup i Got the answer thanks :)
jim_thompson5910
  • jim_thompson5910
ok great
anonymous
  • anonymous
can you please help me with couple of them too?
jim_thompson5910
  • jim_thompson5910
sure a few more
anonymous
  • anonymous
ok here is another one
anonymous
  • anonymous
\[10^{-3x}\times10^{x}=\frac{ 1 }{ 10 }\]
jim_thompson5910
  • jim_thompson5910
\[10^{-3x}\times10^{x}=\frac{ 1 }{ 10 }\] \[10^{-3x}\times10^{x}=10^{-1}\] \[10^{-3x+x}=10^{-1}\] I'll let you finish
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
what do you get
anonymous
  • anonymous
\[x=\frac{ 1 }{ 2 }\]
anonymous
  • anonymous
I need help with couple more pleaseeee!!
anonymous
  • anonymous
\[64\times16^{-3x}=16^{3x-2}\]
jim_thompson5910
  • jim_thompson5910
-3x + x = -1 -2x = -1 x = 1/2 good
anonymous
  • anonymous
thanks
jim_thompson5910
  • jim_thompson5910
\[64\times16^{-3x}=16^{3x-2}\] \[(4^3)\times(4^2)^{-3x}=(4^2)^{3x-2}\] \[(4^3)\times4^{2(-3x)}=(4)^{2(3x-2)}\] \[(4^3)\times4^{-6x}=(4)^{6x-4}\] \[4^{3+(-6x)}=(4)^{6x-4}\] \[4^{3-6x}=(4)^{6x-4}\]
anonymous
  • anonymous
I got an answer: \[x=\frac{ 7 }{ 12 }\]
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • anonymous
\[\frac{ 81^{3n+2} }{ 243^{-n} }=3^{4}\]
anonymous
  • anonymous
oh maybe I can do this one let me try first :)
jim_thompson5910
  • jim_thompson5910
alright, tell me what you get
anonymous
  • anonymous
I got \[-\frac{ 4 }{ 17 }\]
jim_thompson5910
  • jim_thompson5910
you nailed it
anonymous
  • anonymous
:)
anonymous
  • anonymous
\[81\times9^{-2b-2}=27\]
jim_thompson5910
  • jim_thompson5910
tell me what you get
anonymous
  • anonymous
ok
anonymous
  • anonymous
oh I got that one \[-\frac{ 3 }{ 4 }\]
jim_thompson5910
  • jim_thompson5910
good, that's correct
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[\left(\begin{matrix}1 \\ 6\end{matrix}\right)^{3x+2}\times216^{3x}=\frac{ 1 }{ 216 }\]
jim_thompson5910
  • jim_thompson5910
and what did you get
anonymous
  • anonymous
I m having trouble starting it but still trying
jim_thompson5910
  • jim_thompson5910
\[\Large \left(\frac{1}{6}\right)^{3x+2}\times216^{3x}=\frac{ 1 }{ 216 }\] \[\Large \left(6^{-1}\right)^{3x+2}\times(6^3)^{3x}=6^{-3}\] \[\Large \left(6\right)^{-(3x+2)}\times(6)^{9x}=6^{-3}\] \[\Large 6^{-(3x+2)+9x}=6^{-3}\]
anonymous
  • anonymous
oh I only messed up with one sign

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