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ArkGoLucky Group TitleBest ResponseYou've already chosen the best response.0
\[25^{\log_{4}x }5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}25^{\log _{16}x}\]
 one year ago

ArkGoLucky Group TitleBest ResponseYou've already chosen the best response.0
wow that's small
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
\[\Large{25^{\log_{4}x }5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}25^{\log _{16}x}}\]
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
What grade are you?
 one year ago

ArkGoLucky Group TitleBest ResponseYou've already chosen the best response.0
doe that help solve the problem
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
No I just don't see how you can be doing these sort of things in grade 10
 one year ago

ArkGoLucky Group TitleBest ResponseYou've already chosen the best response.0
yeah me neither
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
Okay, I think x is 4, which makes the overall equation 0 = 0.
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
\[\large{25^{\log_{4}4 }5^{(\log _{16}4 ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}25^{\log _{16}4}}\] \[\large{255^{2}=\log _{\sqrt{3}}\sqrt{243}25^{0.5}}\] \[2525 = 0 =\log({\frac{\sqrt{243}}{\sqrt{3}})5} = 5  5 = 0\]
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
How did you realise it had to be a multiple of 4?
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
And when \(x = 4\) gave me \(\log_{4}4\) and \(\log_{16}16\) on the left side, I figured I was on to something,
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
Makes a lot of sense.
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
I did notice one mistake in my explanation, however. It should say: \[\frac{\log{\sqrt{243}}}{\log{\sqrt{3}}}\] not \[\log({\frac{\sqrt{243}}{\sqrt{3}}})\]
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
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. \[\Large{25^{\log_{4}x }5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}25^{\log _{16}x}}\] \[\Large{25^{\log_{4}x }25^{\log _{16}x ^{2}}=525^{\log _{16}x}}\] \[\Large{25^{2 \log_{16}x }25^{\log _{16}x ^{2}}=525^{\log _{16}x}}\] \[\Large{25^{2 \log_{16}x }25^{\log _{16}x ^{2}}+25^{\log _{16}x}=5}\] \[\Large{25^{2 \log_{16}x }25^{2\log _{16}x}+25^{\log _{16}x}=5}\] \[\Large{25^{\log _{16}x}=5}\] \[\Large{5^{2 \log _{16}x}=5}\] \[\Large{2 \log _{16}x=1}\] \[\Large{\log _{16}x=0.5}\] \[\Large{x = 16^{0.5}=\sqrt{16} = 4}\]
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
How did you get to the second step lol
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
\[\Large{\log_{\sqrt{3}}9 \sqrt{3} = 5}\] \[\Large{5^{(\log_{16}x^{2})+1} = 5^{\log_{16}x^{2}} \times 5^{1} = 25^{\log_{16}x^{2}}}\]
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
Ah I didn't think it would come out to be a whole number!! Great job on it.
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
Thanks @Skaematik :)
 one year ago
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