SOLUTION: How do I solve these equations? a) {{{log(6,(x+5))+ log(6,x)=2}}} b) {{{logx=log (2,8)}}} Thank you very much for helping me.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do I solve these equations? a) {{{log(6,(x+5))+ log(6,x)=2}}} b) {{{logx=log (2,8)}}} Thank you very much for helping me.      Log On


   

Question 1043188: How do I solve these equations?
a) log%286%2C%28x%2B5%29%29%2B+log%286%2Cx%29=2
b) logx=log+%282%2C8%29
Thank you very much for helping me.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(a)
+log%286%2C%28x%2B5%29%29%2B+log%286%2Cx%29=2+
The trick here is to express +2+ as a log base 6
+2+=+log%28+6%2C+36+%29+
Now you have:
+log%286%2C%28x%2B5%29%29%2B+log%286%2Cx%29=log%286%2C36%29+
+log%28+6%2C+x%2A%28x%2B5%29+%29+=+log%28+6%2C36+%29+
+x%2A%28+x%2B5+%29+=+36+
+x%5E2+%2B+5x+-+36+=+0+
+%28+x+%2B+9+%29%2A%28+x+-+4+%29+=+0+ ( by looking at it )
+x+=+4+
+x+=+-9+
--------------
+x+=+-9+ won't work as an answer because you get
+log%28+6%2C+x+%29+=+log%28+6%2C+-9+%29+
There is no way a real log to base 6 can give you a negative
number. Even negative logs give you positive answers.
So, +x+=+4+ is the answer
check:
+log%286%2C%28x%2B5%29%29%2B+log%286%2Cx%29=2+
+log%286%2C%284%2B5%29%29%2B+log%286%2C4%29=2+
+log%28+6%2C9+%29+%2B+log%28+6%2C4+%29+=+2+
+log%28+6%2C36+%29+=+2+
+2+=+2+
OK
----------------------------
(b)
+log%28x%29+=+log%282%2C8+%29+
You should know that +log%28+2%2C8+%29+=+3+ so
+log%28x%29+=+3+
The base is assumed to be +10+, so
+x+=+10%5E3+
+x+=+1000+
OK