Solve the logarithmic equation for x: log base 2(x+1)+log base 2(x-1)=3

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.

A community for students.

Solve the logarithmic equation for x: log base 2(x+1)+log base 2(x-1)=3

Mathematics
See more answers at brainly.com
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 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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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...
I mean in class, or where you learned about the logarithm. Or how do you want to solve any problem containing it, if you don't know what it is?
I learned this like 6 months ago, so I really don't remember. This is a take home test... I just need to solve for x, I think.
Then you should look it up, in your notes actually, but wikipedia is often good too. What you need here is \[b^{\log_b x} = x\] because logarithm is the inverse function of the exponential.
Is b the base?
Indeed
So it would be \[2^\log2(3)=3\]?
hi nowhere man, we need to combine the logs first before applying exponentiation. Also in US usually they don't teach exponentiation explicitly in that way.
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 :) )
Formula log(A) to the base 2 + log(B) to the base 2 = log(A B) to the base 2; Steps: 1. log (x+1)(x-1) to the base 2 = 3 2. removing log, 2 raised to 3 = (x+1) (x-1) 3. 8 = x^2 - 1 4. 9 = x^2 5. taking square root on both sides, 3 = x
\[\log_{2} x^2-1\]
that would equal 3, right?
great! So now we have \(\log_2(x^2-1)=3\) Do you remember how to change from log equation into exponential equation?
I think if I make the three on the other side a log as well I could do it, but that doesn't seem 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.
To bee a little pedantic from \(x^2=9\) we have \(x=\pm 3\). But when we plug in back x=-3 into the equation we are in trouble because we can't compute log of negative number. SO x=3 is the only solution.
YAYAYAYYAYAYAYAYYAYAY!<3 Thanks so much guys!
Any time!

Not the answer you are looking for?

Search for more explanations.

Ask your own question