Question

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!

Answer 1

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

Answer 2

you can then write in the equivalent exponential form

Answer 3

\[\log_b(x)=y \implies b^y=x\]

Answer 4

\[3^{4}=9x^2\]

Answer 5

ok and I think you said that above

Answer 6

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

Answer 7

x=3 is right
but there is one more solution

Answer 8

Oh, so it would be 3, -3.

Answer 9

yep

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Okay, that makes sense! Thank you!!

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

I made a mistake above
with the domain of f

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

np

Answer 24

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.

Answer 25

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