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burhan101 Group Title

Solve

  • one year ago
  • one year ago

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  1. burhan101 Group Title
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    |dw:1359089106003:dw|

    • one year ago
  2. burhan101 Group Title
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    i can substitute 2^2 for u for all terms but the last .. idk what to do !

    • one year ago
  3. JayDS Group Title
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    hint: first put them all to the lowest base of 2.

    • one year ago
  4. Azteck Group Title
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    \[\frac{ 1 }{ \sqrt{2} }=2^{-\frac{ 1 }{ 2 }}\]

    • one year ago
  5. JayDS Group Title
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    yep, @Azteck has shown u the harder one, try to do the rest yourself. Post it on here so I/others can help u check it.

    • one year ago
  6. Azteck Group Title
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    @JayDS He only wanted to know the last one. Read what he said above.

    • one year ago
  7. JayDS Group Title
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    oh kk sorry, I didn't quite understand how he phrased it.

    • one year ago
  8. Hero Group Title
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    \[4^2(2^{x - 3}) = 16^{x-2}\] Is that the whole thing @burhan101 ?

    • one year ago
  9. Hero Group Title
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    Because if it is, it works out quite nicely

    • one year ago
  10. burhan101 Group Title
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    no @Hero you are missing the last term

    • one year ago
  11. burhan101 Group Title
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    |dw:1359091740887:dw|

    • one year ago
  12. Hero Group Title
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    \[4^2 \dot\ 2^{x - 3} = \frac{16^{x - 2}}{\sqrt{2}} \\4^2\dot\ 2^x2^{-3} = \frac{16^x16^{-2}}{\sqrt{2}} \\\frac{4^2 \dot\ 2^x}{2^3} = \frac{16^x}{16^2 \sqrt{2}} \\\frac{4^2 \dot\ 16^2 \sqrt{2}}{2^3}= \frac{16^x}{2^x} \\512\sqrt{2} = \left(\frac{16}{2}\right)^x \\512\sqrt{2} = \left(8\right)^x \] You should be able to finish it from there by taking logs of both sides and simplifying

    • one year ago
  13. burhan101 Group Title
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    thank you but this is not the method my teacher wants, she wants us to substitute

    • one year ago
  14. Hero Group Title
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    Substitute what?

    • one year ago
  15. burhan101 Group Title
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    like a variable like let's say use q to represent something

    • one year ago
  16. Hero Group Title
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    Why would she want you to do that? It simplifies quite nicely without it.

    • one year ago
  17. burhan101 Group Title
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    I wish i knew -.- something about knowing a wide variety of methods

    • one year ago
  18. Hero Group Title
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    If you take logs of both sides you get: \[\ln(512 \dot\ \sqrt{2}) = x \ln(8) \\\frac{\ln(512) + \ln(\sqrt{2})}{\ln(8)} = x \\\frac{\ln(2^9) + \ln(2^{1/2})}{\ln(2^3)} = x \\\frac{9\ln(2) + .5\ln(2)}{3\ln(2)} = x \\\frac{9.5 \ln(2)}{3 \ln(2)} = x \\\frac{9.5}{3} = x \]

    • one year ago
  19. Azteck Group Title
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    \[(2^2)^2(2^{2x-3})=(16^{x-2})(\frac{ 1 }{ \sqrt{2} })\] \[(2^4)(2^{2x-3})=(2^{4x-8})(2^{-\frac{ 1 }{ 2 }})\] \[2^{2x-3+4}=2^{4x-8-\frac{ 1 }{ 2 }}\] \[2^{2x+1}=2^{4x-\frac{ 17 }{ 2 }}\] \[2x+1=4x-\frac{ 17 }{ 2 }\] \[4x+2=8x-17\] Solve for x.

    • one year ago
  20. Azteck Group Title
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    @Hero you copied the wrong indice.

    • one year ago
  21. Hero Group Title
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    Good job bro.

    • one year ago
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