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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}\]
wow that's small
\[\Large{25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}}\]

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

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