SOLUTION: How do you solve? 11 - 5log3 (x - 5) = 6 log x = 1 - log(x + 3) and e^6^y = 25 Thanks in advanced.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do you solve? 11 - 5log3 (x - 5) = 6 log x = 1 - log(x + 3) and e^6^y = 25 Thanks in advanced.      Log On


   

Question 111488: How do you solve?
11 - 5log3 (x - 5) = 6
log x = 1 - log(x + 3)
and
e^6^y = 25
Thanks in advanced.
Answer by wgunther(43) About Me  (Show Source):
You can put this solution on YOUR website!
first one, 11+-+5log3+%28x+-+5%29+=+6
+5log_3+%28x+-+5%29+=+5+
+log_3+%28x-5%29+=+1+
At this point you can take both sides to be exponents of 3
+3%5E%28log_3+%28x+-+5%29%29+=+3%5E6
x-5=3%5E6
x=3%5E6%2B5
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second one log+x+=+1+-+log%28x+%2B+3%29
take both sides to be an exponent of 10
10%5E%28log+x%29+=+10%5E%281+-+log%28x+%2B+3%29%29
+x+=+10%2F%28x%2B3%29+
x%5E2+%2B+x+-+10=0
Just check for extranous values, because there is only one solution to the problem but the quadratic will give you two.
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third one. Here, it would have been helpful if you had been more clear. is it %28e%5E6%29%5Ey or e%5E%286%5Ey%29. This is not communitive, meaning these mean different things. I'm going to assume you meant the second one because that is the literal translation of what you said.
e%5E6%5Ey+=+25
Take natural log of both sides
+6%5Ey+=+ln%2825%29+
Take log base 6 of both sides
+y+=+log_6%28ln%2825%29%29