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!

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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!

Mathematics
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\[\log_3(x^2)+\log_3(9)=4 \\ \text{ yes you can use product rule } \\ \log_3(9x^2)=4\]
you can then write in the equivalent exponential form
\[\log_b(x)=y \implies b^y=x\]

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\[3^{4}=9x^2\]
ok and I think you said that above
\[81=9x^2 \\ \frac{81}{9}=x^2 \\ x^2=9\] you should get two solutions not just one
x=3 is right but there is one more solution
Oh, so it would be 3, -3.
yep
Ok, thank you! For some reason, I had the mindset that it couldn't be a negative number. You helped so much!
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
you know since (-3)^2 =9 which is a positive number
your aim is to make sure the solutions you get do not make the inside of the logs negative or zero
Okay, that makes sense! Thank you!!
\[f(x)=\log(x^2) \text{ has domain all real numbers } \\ \text{ while } g(x)=\log(x) \text{ has domain } x>0\]
so if the equation was: \[2 \log_3(x)+\log_3(9)=4 \\ \text{ you would want \to make sure your solutions satisfy } x>0\]
\[\log_3(x^2)+\log_3(9)=4 \text{ by power rule } \\ \log_3(9x^2)=4 \text{ by product rule } \\ 3^{4}=9x^2 \text{ equivalent exponential form } \\ 81=9x^2 \\ x^2=9 \\ x=\pm 3 \\ \text{ but we said } x>0 \text{ so only } x=3 \\ \text{ works for } 2 \log_3(x)+\log_3(9)=4\]
I made a mistake above with the domain of f
\[f(x)=\log(x^2) \text{ has domain all real numbers excluding } x=0\]
anyways I hope you see I was solving a slightly different equation right there
\[2 \log_3(x)+\log_3(9)=4 \text{ has solutions } x=3 \\ \log_3(x^2)+\log_3(9) =4 \text{ has solutions } x=3,-3\]
Okay, you have made this a lot easier! The pieces of information finally fit now!! Thank you so much!
np
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.
Thanks! I'm pretty new to this site, and I will definitely stick with it because the help is awesome!

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