SOLUTION: Please help me answer the following! And a bit side info would help.. A. Expand 1) ln (x^2 y)/z 2) 4th root of (x^2 y)/z^3 3) log to the base 5 (x-7)/sqr. Root of x-1(2+x)^4

Question 821502: Please help me answer the following! And a bit side info would help..
A. Expand
1) ln (x^2 y)/z
2) 4th root of (x^2 y)/z^3
3) log to the base 5 (x-7)/sqr. Root of x-1(2+x)^4
B. Condense
1) 2(log sqr. Root of x-5 + 3 log 6th root of x+5)
2) 1 + log of 12 to the base 3 - 1/2log of 8 to the base 3 - logof 2 to the base 9
3) 4-ln4+ln6
Thank you veeery much in advance!

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A. Expand
1) ln [(x^2 y)/z] = 2ln(x) + ln(y) - ln(z)
--------------------------------
2) 4th root of (x^2 y)/z^3 = x^(1/2)*y^(1/4)/z^(3/4)
--------------------------------
3) log to the base 5 (x-7)/sqr. Root of x-1(2+x)^4
= log5[(x-7)/sqrt(x-(2+x)^4)]
= log5(x-7) - (1/2)log5[x-(2+x)^4]
Note: I doubt that is correct. Your post is impossible to interpret.
--------------------------------------------------------
B. Condense
1) 2(log sqr. Root of x-5 + 3 log 6th root of x+5)
=====
= 2*(log(sqrt(x-5) + 3log[(x+5)^(1/6)])
= 2*(log[sqrt(x-5)*(x+5)^(1/2)]
-------------
2) 1 + log of 12 to the base 3 - 1/2log of 8 to the base 3 - logof 2 to the base 9
-----
= 1 + log3(12) - log3(8^(1/2)) - log9(2)
-----
= 1 + log3[12/2sqrt(2)] - 2log3(2)
= 1 + log3[12/[2sqrt(2)*4]
= 1 + log3[(3/2sqrt(2)]
= 1 + log3(3)- log3(2sqrt(2)
-----
= 2 - log3(2sqrt(2))
-------
3) 4-ln4+ln6
---
= 4 + ln(6/4)
----
= 4 + ln(3/2)
===================
Cheers,
Stan H.
===================

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