## Meehan98 one year ago So, I need some help with Logarithms. The equation is: log (subscript 3) x^2 + log (subscript 3) 9 = 4. I thought you multiply (x^2)(9) in order to get one log on the left side, so you're left with... log (subscript 3) 9x^2=4 and because the log is the exponent, you have log (subscript 3) 9x^2=3^4=81, which makes x=3, but that is the wrong answer. Thank You!

1. freckles

$\log_3(x^2)+\log_3(9)=4 \\ \text{ yes you can use product rule } \\ \log_3(9x^2)=4$

2. freckles

you can then write in the equivalent exponential form

3. freckles

$\log_b(x)=y \implies b^y=x$

4. freckles

$3^{4}=9x^2$

5. freckles

ok and I think you said that above

6. freckles

$81=9x^2 \\ \frac{81}{9}=x^2 \\ x^2=9$ you should get two solutions not just one

7. freckles

x=3 is right but there is one more solution

8. Meehan98

Oh, so it would be 3, -3.

9. freckles

yep

10. Meehan98

Ok, thank you! For some reason, I had the mindset that it couldn't be a negative number. You helped so much!

11. freckles

when you plug in -3 into the equation you have that one term is log((-3)^2)) the inside of that log is positive so it is cool

12. freckles

you know since (-3)^2 =9 which is a positive number

13. freckles

your aim is to make sure the solutions you get do not make the inside of the logs negative or zero

14. Meehan98

Okay, that makes sense! Thank you!!

15. freckles

$f(x)=\log(x^2) \text{ has domain all real numbers } \\ \text{ while } g(x)=\log(x) \text{ has domain } x>0$

16. freckles

so if the equation was: $2 \log_3(x)+\log_3(9)=4 \\ \text{ you would want \to make sure your solutions satisfy } x>0$

17. freckles

$\log_3(x^2)+\log_3(9)=4 \text{ by power rule } \\ \log_3(9x^2)=4 \text{ by product rule } \\ 3^{4}=9x^2 \text{ equivalent exponential form } \\ 81=9x^2 \\ x^2=9 \\ x=\pm 3 \\ \text{ but we said } x>0 \text{ so only } x=3 \\ \text{ works for } 2 \log_3(x)+\log_3(9)=4$

18. freckles

I made a mistake above with the domain of f

19. freckles

$f(x)=\log(x^2) \text{ has domain all real numbers excluding } x=0$

20. freckles

anyways I hope you see I was solving a slightly different equation right there

21. freckles

$2 \log_3(x)+\log_3(9)=4 \text{ has solutions } x=3 \\ \log_3(x^2)+\log_3(9) =4 \text{ has solutions } x=3,-3$

22. Meehan98

Okay, you have made this a lot easier! The pieces of information finally fit now!! Thank you so much!

23. freckles

np

24. freckles

I gave you medal because you are awesome. You showed your work. And that totally rules in my opinion. You also did pretty well on solving it.

25. Meehan98

Thanks! I'm pretty new to this site, and I will definitely stick with it because the help is awesome!