Meehan98
  • Meehan98
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
\[\log_3(x^2)+\log_3(9)=4 \\ \text{ yes you can use product rule } \\ \log_3(9x^2)=4\]
freckles
  • freckles
you can then write in the equivalent exponential form
freckles
  • freckles
\[\log_b(x)=y \implies b^y=x\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
\[3^{4}=9x^2\]
freckles
  • freckles
ok and I think you said that above
freckles
  • freckles
\[81=9x^2 \\ \frac{81}{9}=x^2 \\ x^2=9\] you should get two solutions not just one
freckles
  • freckles
x=3 is right but there is one more solution
Meehan98
  • Meehan98
Oh, so it would be 3, -3.
freckles
  • freckles
yep
Meehan98
  • Meehan98
Ok, thank you! For some reason, I had the mindset that it couldn't be a negative number. You helped so much!
freckles
  • 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
freckles
  • freckles
you know since (-3)^2 =9 which is a positive number
freckles
  • freckles
your aim is to make sure the solutions you get do not make the inside of the logs negative or zero
Meehan98
  • Meehan98
Okay, that makes sense! Thank you!!
freckles
  • freckles
\[f(x)=\log(x^2) \text{ has domain all real numbers } \\ \text{ while } g(x)=\log(x) \text{ has domain } x>0\]
freckles
  • freckles
so if the equation was: \[2 \log_3(x)+\log_3(9)=4 \\ \text{ you would want \to make sure your solutions satisfy } x>0\]
freckles
  • freckles
\[\log_3(x^2)+\log_3(9)=4 \text{ by power rule } \\ \log_3(9x^2)=4 \text{ by product rule } \\ 3^{4}=9x^2 \text{ equivalent exponential form } \\ 81=9x^2 \\ x^2=9 \\ x=\pm 3 \\ \text{ but we said } x>0 \text{ so only } x=3 \\ \text{ works for } 2 \log_3(x)+\log_3(9)=4\]
freckles
  • freckles
I made a mistake above with the domain of f
freckles
  • freckles
\[f(x)=\log(x^2) \text{ has domain all real numbers excluding } x=0\]
freckles
  • freckles
anyways I hope you see I was solving a slightly different equation right there
freckles
  • freckles
\[2 \log_3(x)+\log_3(9)=4 \text{ has solutions } x=3 \\ \log_3(x^2)+\log_3(9) =4 \text{ has solutions } x=3,-3\]
Meehan98
  • Meehan98
Okay, you have made this a lot easier! The pieces of information finally fit now!! Thank you so much!
freckles
  • freckles
np
freckles
  • 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.
Meehan98
  • Meehan98
Thanks! I'm pretty new to this site, and I will definitely stick with it because the help is awesome!

Looking for something else?

Not the answer you are looking for? Search for more explanations.