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Pinky1234

  • 3 years ago

Help Please. 4^2x+3=1

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  1. Pinky1234
    • 3 years ago
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    I need help badly in couple of questions, someone please help!!

  2. jim_thompson5910
    • 3 years ago
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    is the problem \[\Large 4^{2x}+3=1\] OR is it \[\Large 4^{2x+3}=1\]

  3. Pinky1234
    • 3 years ago
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    yeah that second one

  4. jim_thompson5910
    • 3 years ago
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    Hint: \[\Large 4^{2x+3}=1\] \[\Large 4^{2x+3}=4^0\]

  5. Pinky1234
    • 3 years ago
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    so we are not using that 1?

  6. jim_thompson5910
    • 3 years ago
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    we are, just a different form of it 4^0 is the same as 1

  7. Pinky1234
    • 3 years ago
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    ok

  8. jim_thompson5910
    • 3 years ago
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    the bases are the same (4) so the exponents must be the same

  9. jim_thompson5910
    • 3 years ago
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    so 2x + 3 = 0

  10. Pinky1234
    • 3 years ago
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    yup i Got the answer thanks :)

  11. jim_thompson5910
    • 3 years ago
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    ok great

  12. Pinky1234
    • 3 years ago
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    can you please help me with couple of them too?

  13. jim_thompson5910
    • 3 years ago
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    sure a few more

  14. Pinky1234
    • 3 years ago
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    ok here is another one

  15. Pinky1234
    • 3 years ago
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    \[10^{-3x}\times10^{x}=\frac{ 1 }{ 10 }\]

  16. jim_thompson5910
    • 3 years ago
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    \[10^{-3x}\times10^{x}=\frac{ 1 }{ 10 }\] \[10^{-3x}\times10^{x}=10^{-1}\] \[10^{-3x+x}=10^{-1}\] I'll let you finish

  17. Pinky1234
    • 3 years ago
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    ok

  18. jim_thompson5910
    • 3 years ago
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    what do you get

  19. Pinky1234
    • 3 years ago
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    \[x=\frac{ 1 }{ 2 }\]

  20. Pinky1234
    • 3 years ago
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    I need help with couple more pleaseeee!!

  21. Pinky1234
    • 3 years ago
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    \[64\times16^{-3x}=16^{3x-2}\]

  22. jim_thompson5910
    • 3 years ago
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    -3x + x = -1 -2x = -1 x = 1/2 good

  23. Pinky1234
    • 3 years ago
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    thanks

  24. jim_thompson5910
    • 3 years ago
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    \[64\times16^{-3x}=16^{3x-2}\] \[(4^3)\times(4^2)^{-3x}=(4^2)^{3x-2}\] \[(4^3)\times4^{2(-3x)}=(4)^{2(3x-2)}\] \[(4^3)\times4^{-6x}=(4)^{6x-4}\] \[4^{3+(-6x)}=(4)^{6x-4}\] \[4^{3-6x}=(4)^{6x-4}\]

  25. Pinky1234
    • 3 years ago
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    I got an answer: \[x=\frac{ 7 }{ 12 }\]

  26. jim_thompson5910
    • 3 years ago
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    good

  27. Pinky1234
    • 3 years ago
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    \[\frac{ 81^{3n+2} }{ 243^{-n} }=3^{4}\]

  28. Pinky1234
    • 3 years ago
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    oh maybe I can do this one let me try first :)

  29. jim_thompson5910
    • 3 years ago
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    alright, tell me what you get

  30. Pinky1234
    • 3 years ago
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    I got \[-\frac{ 4 }{ 17 }\]

  31. jim_thompson5910
    • 3 years ago
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    you nailed it

  32. Pinky1234
    • 3 years ago
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    :)

  33. Pinky1234
    • 3 years ago
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    \[81\times9^{-2b-2}=27\]

  34. jim_thompson5910
    • 3 years ago
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    tell me what you get

  35. Pinky1234
    • 3 years ago
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    ok

  36. Pinky1234
    • 3 years ago
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    oh I got that one \[-\frac{ 3 }{ 4 }\]

  37. jim_thompson5910
    • 3 years ago
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    good, that's correct

  38. Pinky1234
    • 3 years ago
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    ok

  39. Pinky1234
    • 3 years ago
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    \[\left(\begin{matrix}1 \\ 6\end{matrix}\right)^{3x+2}\times216^{3x}=\frac{ 1 }{ 216 }\]

  40. jim_thompson5910
    • 3 years ago
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    and what did you get

  41. Pinky1234
    • 3 years ago
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    I m having trouble starting it but still trying

  42. jim_thompson5910
    • 3 years ago
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    \[\Large \left(\frac{1}{6}\right)^{3x+2}\times216^{3x}=\frac{ 1 }{ 216 }\] \[\Large \left(6^{-1}\right)^{3x+2}\times(6^3)^{3x}=6^{-3}\] \[\Large \left(6\right)^{-(3x+2)}\times(6)^{9x}=6^{-3}\] \[\Large 6^{-(3x+2)+9x}=6^{-3}\]

  43. Pinky1234
    • 3 years ago
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    oh I only messed up with one sign

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