SOLUTION: Express the following in index form Logx + logy =1 Logx =6 Log (x+4) = log64 Log (2x+1) - log (3x-2) Log (3x-1) - log2 = 3

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Express the following in index form Logx + logy =1 Logx =6 Log (x+4) = log64 Log (2x+1) - log (3x-2) Log (3x-1) - log2 = 3      Log On


   

Question 944015: Express the following in index form
Logx + logy =1
Logx =6
Log (x+4) = log64
Log (2x+1) - log (3x-2)
Log (3x-1) - log2 = 3
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Express the following in index form
Logx + logy =1
log(xy) = 1
xy = 10
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Logx =6
x = 10^6 = 1000000
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Log (x+4) = log64
log(x+4)/(64) = 0
(x+4)/64 = 1
x+4 = 64
x = 60
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Log (2x+1) - log (3x-2)
---
= log[(2x+1)/(3x-2)]
=======================
Log (3x-1) - log2 = 3
----
log[(3x-1)/2] = 3
----
(3x-1)/2 = 10^3
-----
3x-1 = 2000
3x = 2001
x = 667
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cheers,
Stan H.
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