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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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

you can then write in the equivalent exponential form

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

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

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

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!!

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

np