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ArkGoLucky

  • 3 years ago

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  1. ArkGoLucky
    • 3 years ago
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    \[25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}\]

  2. ArkGoLucky
    • 3 years ago
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    wow that's small

  3. geoffb
    • 3 years ago
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    \[\Large{25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}}\]

  4. ArkGoLucky
    • 3 years ago
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    yes

  5. Skaematik
    • 3 years ago
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    What grade are you?

  6. ArkGoLucky
    • 3 years ago
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    10th

  7. ArkGoLucky
    • 3 years ago
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    doe that help solve the problem

  8. Skaematik
    • 3 years ago
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    No I just don't see how you can be doing these sort of things in grade 10

  9. ArkGoLucky
    • 3 years ago
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    yeah me neither

  10. geoffb
    • 3 years ago
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    I wouldn't know where to start for this, but I can guess your answer will be a multiple of 4. If not, it could get ugly. :P

  11. geoffb
    • 3 years ago
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    Okay, I think x is 4, which makes the overall equation 0 = 0.

  12. geoffb
    • 3 years ago
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    \[\large{25^{\log_{4}4 }-5^{(\log _{16}4 ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}4}}\] \[\large{25-5^{2}=\log _{\sqrt{3}}\sqrt{243}-25^{0.5}}\] \[25-25 = 0 =\log({\frac{\sqrt{243}}{\sqrt{3}})-5} = 5 - 5 = 0\]

  13. Skaematik
    • 3 years ago
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    How did you realise it had to be a multiple of 4?

  14. geoffb
    • 3 years ago
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    Because all the bases of the logs were multiples of 4 (either 4 or 16), so I knew anything that wasn't a multiple of 4 would be too messy.

  15. geoffb
    • 3 years ago
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    And when \(x = 4\) gave me \(\log_{4}4\) and \(\log_{16}16\) on the left side, I figured I was on to something,

  16. Skaematik
    • 3 years ago
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    Makes a lot of sense.

  17. geoffb
    • 3 years ago
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    I did notice one mistake in my explanation, however. It should say: \[\frac{\log{\sqrt{243}}}{\log{\sqrt{3}}}\] not \[\log({\frac{\sqrt{243}}{\sqrt{3}}})\]

  18. geoffb
    • 3 years ago
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    Okay, I've tried to include as many parts of my process as I could. If you have any questions about what I did, @ArkGoLucky, let me know. \[\Large{25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}}\] \[\Large{25^{\log_{4}x }-25^{\log _{16}x ^{2}}=5-25^{\log _{16}x}}\] \[\Large{25^{2 \log_{16}x }-25^{\log _{16}x ^{2}}=5-25^{\log _{16}x}}\] \[\Large{25^{2 \log_{16}x }-25^{\log _{16}x ^{2}}+25^{\log _{16}x}=5}\] \[\Large{25^{2 \log_{16}x }-25^{2\log _{16}x}+25^{\log _{16}x}=5}\] \[\Large{25^{\log _{16}x}=5}\] \[\Large{5^{2 \log _{16}x}=5}\] \[\Large{2 \log _{16}x=1}\] \[\Large{\log _{16}x=0.5}\] \[\Large{x = 16^{0.5}=\sqrt{16} = 4}\]

  19. Skaematik
    • 3 years ago
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    How did you get to the second step lol

  20. geoffb
    • 3 years ago
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    \[\Large{\log_{\sqrt{3}}9 \sqrt{3} = 5}\] \[\Large{5^{(\log_{16}x^{2})+1} = 5^{\log_{16}x^{2}} \times 5^{1} = 25^{\log_{16}x^{2}}}\]

  21. Skaematik
    • 3 years ago
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    Ah I didn't think it would come out to be a whole number!! Great job on it.

  22. geoffb
    • 3 years ago
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    Thanks @Skaematik :)

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