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!

So how far do you get?

not at all because its so confusing

is this log base 5?

actually it doesnt matter so lets just write log

and ignore the 1/3 out front we will deal with it last

yes its a base 5

first use
\[\log(a^n)=n\log(a)\] backwards. all the multipliers come inside the log as exponents

\[4\log(x+1)=\log((x+1)^4)\]
\[3\log(3-x)=\log((3-x)^3)\]
\[6\log(x)=\log(x^6)\]

giving
\[\log((x+1)^4)-\log((3-x)^3)+\log(x^6)\]

now we use \[\log(a)-log(b)=log(\frac{a}{b})\]

what happened to the 1/3 outside the []

i will deal with that last

not done yet!

\[log((x+1)^4)-\log((3-x)^3)=\log(\frac{(x+1)^4}{(3-x)^3})\]

now we use
\[\log(a)+\log(b)=\log(ab)\]

\[\log(\frac{(x+1)^4}{(3-x)^3})+\log(x^6)=\log(\frac{x^6(x+1)^4}{(3-x)^3})\]

now for the 1/3 out front. take the cube root of all that stuff.

wow that was a hard problem but thanks again:)