SOLUTION: Write as a single log expression:
2(log(x)+log(x^2+1))-3log(x-2)
log(x^2)+log(x^4+1)-log(x^3-8)
log(x^2(x^4+1)-log(x^3-8)
log(x^6+x^2)-log(x^3-8)
log(x^12)-log(x^3-8)
log[(x^
Algebra.Com
Question 376752: Write as a single log expression:
2(log(x)+log(x^2+1))-3log(x-2)
log(x^2)+log(x^4+1)-log(x^3-8)
log(x^2(x^4+1)-log(x^3-8)
log(x^6+x^2)-log(x^3-8)
log(x^12)-log(x^3-8)
log[(x^12)/(x^3-8)]
???
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The regular rules of algebra don't get suspended just because you are working with logarithms.
\ +\ \log\left(x^2\ +\ 1\right)\right)\ -\ 3\log(x\ -\ 2))
First use
 + \log_b(\upsilon)\ =\ \log_b(\chi\upsilon))
to write:
\right)\ -\ 3\log(x\ -\ 2))
Then use
\ =\ \log_b(\chi^n))
To write:
^2\right)\ -\ \log\left((x\ -\ 2)^3\right))
Then use
 - \log_b(\upsilon)\ =\ log_b\left(\frac{\chi}{\upsilon}\right))
To write:
^2}{(x\ -\ 2)^3}\right))
Which satisfies the requirements of the problem. You could multiply out the two binomials, but that would be a decision based on the requirements of further calculations.
John
My calculator said it, I believe it, that settles it
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