SOLUTION: Hey hows it goin I ran into a little issue with log's I was wondering if you could help me solve this one. The base of the log's are a understood 10. log4x=log5+log(x-4)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hey hows it goin I ran into a little issue with log's I was wondering if you could help me solve this one. The base of the log's are a understood 10. log4x=log5+log(x-4)      Log On


   

Question 423152: Hey hows it goin I ran into a little issue with log's I was wondering if you could help me solve this one. The base of the log's are a understood 10.
log4x=log5+log(x-4)
Found 2 solutions by jim_thompson5910, josmiceli:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
log4x=log5+log(x-4)

log4x=log[5(x-4)]

4x=5(x-4)

4x=5x-20

4x-5x=-20

-x=-20

x=20


If you need more help, email me at jim_thompson5910@hotmail.com

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Jim

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+log%284x%29+=+log%285%29+%2B+log%28x-4%29+
+log%284x%29+=+log+%285%2A%28x-4%29%29+
The logs are equal, so
+4x+=+5x+-+20+
+x+=+20+
check answer:
+log%284x%29=log%285%29%2Blog%28x-4%29+
+log%2880%29+=+log%285%29+%2B+log%2820+-+4%29+
+log%2880%29+=+log%285%2A16%29+
+log%2880%29+=+log%2880%29+
OK