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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!
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!

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\log_3(x^2)+\log_3(9)=4 \\ \text{ yes you can use product rule } \\ \log_3(9x^2)=4\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you can then write in the equivalent exponential form

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\log_b(x)=y \implies b^y=x\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok and I think you said that above

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[81=9x^2 \\ \frac{81}{9}=x^2 \\ x^2=9\] you should get two solutions not just one

freckles
 one year ago
Best ResponseYou've already chosen the best response.1x=3 is right but there is one more solution

Meehan98
 one year ago
Best ResponseYou've already chosen the best response.1Oh, so it would be 3, 3.

Meehan98
 one year ago
Best ResponseYou've already chosen the best response.1Ok, thank you! For some reason, I had the mindset that it couldn't be a negative number. You helped so much!

freckles
 one year ago
Best ResponseYou've already chosen the best response.1when 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you know since (3)^2 =9 which is a positive number

freckles
 one year ago
Best ResponseYou've already chosen the best response.1your aim is to make sure the solutions you get do not make the inside of the logs negative or zero

Meehan98
 one year ago
Best ResponseYou've already chosen the best response.1Okay, that makes sense! Thank you!!

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x)=\log(x^2) \text{ has domain all real numbers } \\ \text{ while } g(x)=\log(x) \text{ has domain } x>0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so if the equation was: \[2 \log_3(x)+\log_3(9)=4 \\ \text{ you would want \to make sure your solutions satisfy } x>0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\log_3(x^2)+\log_3(9)=4 \text{ by power rule } \\ \log_3(9x^2)=4 \text{ by product rule } \\ 3^{4}=9x^2 \text{ equivalent exponential form } \\ 81=9x^2 \\ x^2=9 \\ x=\pm 3 \\ \text{ but we said } x>0 \text{ so only } x=3 \\ \text{ works for } 2 \log_3(x)+\log_3(9)=4\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I made a mistake above with the domain of f

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x)=\log(x^2) \text{ has domain all real numbers excluding } x=0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1anyways I hope you see I was solving a slightly different equation right there

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[2 \log_3(x)+\log_3(9)=4 \text{ has solutions } x=3 \\ \log_3(x^2)+\log_3(9) =4 \text{ has solutions } x=3,3\]

Meehan98
 one year ago
Best ResponseYou've already chosen the best response.1Okay, you have made this a lot easier! The pieces of information finally fit now!! Thank you so much!

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

Meehan98
 one year ago
Best ResponseYou've already chosen the best response.1Thanks! I'm pretty new to this site, and I will definitely stick with it because the help is awesome!
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