Question

Log question

Answer 1

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

Answer 2

wow that's small

Answer 3

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

Answer 4

yes

Answer 5

What grade are you?

Answer 6

10th

Answer 7

doe that help solve the problem

Answer 8

No I just don't see how you can be doing these sort of things in grade 10

Answer 9

yeah me neither

Answer 10

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

Answer 11

Okay, I think x is 4, which makes the overall equation 0 = 0.

Answer 12

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

Answer 13

How did you realise it had to be a multiple of 4?

Answer 14

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.

Answer 15

And when \(x = 4\) gave me \(\log_{4}4\) and \(\log_{16}16\) on the left side, I figured I was on to something,

Answer 16

Makes a lot of sense.

Answer 17

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

Answer 18

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.
\[\Large{25^{\log_{4}x }-5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}-25^{\log _{16}x}}\]
\[\Large{25^{\log_{4}x }-25^{\log _{16}x ^{2}}=5-25^{\log _{16}x}}\]
\[\Large{25^{2 \log_{16}x }-25^{\log _{16}x ^{2}}=5-25^{\log _{16}x}}\]
\[\Large{25^{2 \log_{16}x }-25^{\log _{16}x ^{2}}+25^{\log _{16}x}=5}\]
\[\Large{25^{2 \log_{16}x }-25^{2\log _{16}x}+25^{\log _{16}x}=5}\]
\[\Large{25^{\log _{16}x}=5}\]
\[\Large{5^{2 \log _{16}x}=5}\]
\[\Large{2 \log _{16}x=1}\]
\[\Large{\log _{16}x=0.5}\]
\[\Large{x = 16^{0.5}=\sqrt{16} = 4}\]

Answer 19

How did you get to the second step lol

Answer 20

\[\Large{\log_{\sqrt{3}}9 \sqrt{3} = 5}\]
\[\Large{5^{(\log_{16}x^{2})+1} = 5^{\log_{16}x^{2}} \times 5^{1} = 25^{\log_{16}x^{2}}}\]

Answer 21

Ah I didn't think it would come out to be a whole number!! Great job on it.

Answer 22

Thanks @Skaematik :)