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.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

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

Apply exponentiation!

.........huh? :P

If you know the logarithm you should know the exponential.

I'm lost........ Gah why does math have to be so hard?

It's not, if you follow the definitions. How did you define log?

What do you mean? I didn't define anything... I don't think...

Is b the base?

Indeed

So it would be \[2^\log2(3)=3\]?

That's what I thought, cause I didn't remember anything like that.... Help?

Do you remember this property?
\(log_b A+\log_b B= \log_b(AB)?\)

I think so...

Ok, what do you get if you apply that property to the left hand side of your equation?

\[\log_{2} 2x\]I think. Because the +1 and the -1 would cancel out.

But you multiply the inside instead of adding them.

Why would I multiply them? It says I'm adding the two logs...

Answer x = 3;
log(3+1) to the base 2 = 2
and log(3-1) to the base 2 = 1

look at again \(log_b A +\log_b B=\log_b(AB)\)
In there you multiply A and B

Ohhhhh okay. So it would be\[\log_{2} x^2+1\]

Check it again what do you get if you do (x+1)(x-1) (foil it :) )

\[\log_{2} x^2-1\]

that would equal 3, right?

Remember this
\[\log_bx=y\iff b^y=x\]

That's right

OHH I REMEMBER THAT! :D

So it would be \[2^3=x^2-1\]

Yep. Now solve for x :)

\[x^2=9\]so x=3 :D

AHHH YOU GUYS ARE AMAZING<3

x could be -3 as well. Since we are dealing with logarithms, we ignore -3

Oh yeah, I forgot about that. Negatives don't work with logs.

YAYAYAYYAYAYAYAYYAYAY!<3 Thanks so much guys!

Any time!