## anonymous 3 years ago Log question

1. anonymous

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

2. anonymous

wow that's small

3. anonymous

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

4. anonymous

yes

5. anonymous

6. anonymous

10th

7. anonymous

doe that help solve the problem

8. anonymous

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

9. anonymous

yeah me neither

10. 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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. 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.

15. 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,

16. anonymous

Makes a lot of sense.

17. anonymous

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

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

19. anonymous

How did you get to the second step lol

20. anonymous

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

21. anonymous

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

22. anonymous

Thanks @Skaematik :)