SOLUTION: Given that log (base 7)x^2 = p and log (base 7) xy^2 = q, find log (base 7) (cube root over xy) in terms of p and q. hope you'll get my question right!

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Given that log (base 7)x^2 = p and log (base 7) xy^2 = q, find log (base 7) (cube root over xy) in terms of p and q. hope you'll get my question right!      Log On


   

Question 824077: Given that log (base 7)x^2 = p and log (base 7) xy^2 = q, find log (base 7) (cube root over xy) in terms of p and q.
hope you'll get my question right!
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+log%28+7%2C+x%5E2+%29+=+p+
I can say:
+7%5Ep+=+x%5E2+
+x+=+%28+7%5Ep+%29%5E%281%2F2%29+
(1) +x+=+7%5E%28p%2F2%29+
----------------
+log%28+7%2C+x%2Ay%5E2+%29+=+q+
I can say:
+7%5Eq+=+x%2Ay%5E2+
+y%5E2+=+%28+7%5Eq+%29+%2F+x+
(2) +y+=+%28+7%5E%28q%2F2%29%29+%2F+x%5E%281%2F2%29+
----------------------------
From (1),
+x%5E%281%2F2%29+=+7%5E%28p%2F4%29+
---------------------
(2) +y+=+%28+7%5E%28q%2F2%29%29+%2F+7%5E%28p%2F4%29+
----------------------------
+x%2Ay+=+%28+7%5E%28p%2F2%29+%29%2A%28+7%5E%28q%2F2%29+%29+%2F+7%5E%28+p%2F4%29++
+x%2Ay+=++%28+7%5E%282p%2F4%29+%29%2A%28+7%5E%28q%2F2%29+%29+%2F+7%5E%28+p%2F4%29++
+x%2Ay+=++%28+7%5E%28+p%2F4%29+%29%2A%28+7%5E%28q%2F2%29+%29+
+x%2Ay+=+7%5E%28+p%2F4+%2B+q%2F2+%29+
+%28+x%2Ay+%29%5E%281%2F3%29+=+7%5E%28+p%2F12+%2B+q%2F6+%29+
--------------------------------
Rewriting this:
+log%28+7%2C+%28x%2Ay%29%5E%281%2F3%29+%29+=+p%2F12+%2B+q%2F6+
Hope I got it


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Given that log (base 7)x^2 = p and log (base 7) xy^2 = q, find log (base 7) (cube root over xy) in terms of p and q.
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What do you mean by "cube root over xy"?
Cheers,
Stan H.
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