anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}\]
anonymous
  • anonymous
wow that's small
anonymous
  • anonymous
\[\Large{25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}}\]

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anonymous
  • anonymous
yes
anonymous
  • anonymous
What grade are you?
anonymous
  • anonymous
10th
anonymous
  • anonymous
doe that help solve the problem
anonymous
  • anonymous
No I just don't see how you can be doing these sort of things in grade 10
anonymous
  • anonymous
yeah me neither
anonymous
  • anonymous
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
anonymous
  • anonymous
Okay, I think x is 4, which makes the overall equation 0 = 0.
anonymous
  • anonymous
\[\large{25^{\log_{4}4 }-5^{(\log _{16}4 ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}4}}\] \[\large{25-5^{2}=\log _{\sqrt{3}}\sqrt{243}-25^{0.5}}\] \[25-25 = 0 =\log({\frac{\sqrt{243}}{\sqrt{3}})-5} = 5 - 5 = 0\]
anonymous
  • anonymous
How did you realise it had to be a multiple of 4?
anonymous
  • anonymous
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.
anonymous
  • anonymous
And when \(x = 4\) gave me \(\log_{4}4\) and \(\log_{16}16\) on the left side, I figured I was on to something,
anonymous
  • anonymous
Makes a lot of sense.
anonymous
  • anonymous
I did notice one mistake in my explanation, however. It should say: \[\frac{\log{\sqrt{243}}}{\log{\sqrt{3}}}\] not \[\log({\frac{\sqrt{243}}{\sqrt{3}}})\]
anonymous
  • anonymous
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. \[\Large{25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}}\] \[\Large{25^{\log_{4}x }-25^{\log _{16}x ^{2}}=5-25^{\log _{16}x}}\] \[\Large{25^{2 \log_{16}x }-25^{\log _{16}x ^{2}}=5-25^{\log _{16}x}}\] \[\Large{25^{2 \log_{16}x }-25^{\log _{16}x ^{2}}+25^{\log _{16}x}=5}\] \[\Large{25^{2 \log_{16}x }-25^{2\log _{16}x}+25^{\log _{16}x}=5}\] \[\Large{25^{\log _{16}x}=5}\] \[\Large{5^{2 \log _{16}x}=5}\] \[\Large{2 \log _{16}x=1}\] \[\Large{\log _{16}x=0.5}\] \[\Large{x = 16^{0.5}=\sqrt{16} = 4}\]
anonymous
  • anonymous
How did you get to the second step lol
anonymous
  • anonymous
\[\Large{\log_{\sqrt{3}}9 \sqrt{3} = 5}\] \[\Large{5^{(\log_{16}x^{2})+1} = 5^{\log_{16}x^{2}} \times 5^{1} = 25^{\log_{16}x^{2}}}\]
anonymous
  • anonymous
Ah I didn't think it would come out to be a whole number!! Great job on it.
anonymous
  • anonymous
Thanks @Skaematik :)

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