SOLUTION: How do I solve the equation {{{ log2x=4-3logx2 }}} ?

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Question 939657: How do I solve the equation ++log2x=4-3logx2++ ?
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Convert +4+ to a log
+4+=+log%28+10000+%29+
+log%28+2x+%29+=+log%28+10000+%29+-+3%2Alog%28+x%5E2+%29+
( Is this the equation? Hope so )
+log%28+2x+%29+%2B+3%2Alog%28+x%5E2+%29+=+log%28+10000+%29+
+log%28+2x+%29+%2B+log%28+x%5E6+%29+=+log%28+10%5E4+%29+
+log%28+%28+x%5E6+%2A+2x+%29+%29+=+log%28+10%5E4+%29+
+2%2Ax%5E7+=+10%5E4+
+x%5E7+=+%281%2F2%29%2A10%5E4+
+x+=+10%5E%284%2F7%29+%2F+2%5E7+


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log2x=4-3logx2
------------
log(2x) = 4 - log(x^2)^3
----
log(2x) + log(x^6) = 4
----
log[2x*x^6] = 4
----
log[2x^7] = 4
---
2x^7 = 10^4
-----
x^7 = 10000/2 = 5000
-----
x = 5000^(1/7)
------
x = 3.376..
Cheers,
Stan H.
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