SOLUTION: given log2=.3010, log7= .845, and log3=.4771 find the values of: 1)log28 2)log 144/49 please show me how to solve this.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: given log2=.3010, log7= .845, and log3=.4771 find the values of: 1)log28 2)log 144/49 please show me how to solve this.      Log On


   

Question 770458: given log2=.3010, log7= .845, and log3=.4771 find the values of:
1)log28
2)log 144/49
please show me how to solve this.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+log%28+28+%29+=+log%28%28+3%2A7+%29%29+
+log%28%28+3%2A7+%29%29+=+log%28+3+%29+%2B+log%28+7+%29+
+log%28+3+%29+%2B+log%28+7+%29+=+.4771+%2B+.845+
---------------------------------
+log%28+144+%2F+49+%29+=+log%28+12%5E2+%2F+7%5E2+%29+
+log%28+12%5E2+%2F+7%5E2+%29+=+log%28+%28+12%2F7+%29%5E2+%29+
+log%28+%28+12%2F7+%29%5E2+%29+=+2%2Alog%28+12%2F7+%29+
+2%2Alog%28+12%2F7+%29+=+2%2A%28+log%28+12+%29+-+log%28+7+%29+%29+
and
+log%28%28+12+%29%29+=+log%28%28+3%2A2%2A2+%29%29+
+log%28%28+3%2A2%2A2+%29%29+=+log%283%29+%2B+2%2Alog%282%29+
+log%283%29+%2B+2%2Alog%282%29+=+.4771+%2B+2%2A.301+
therefore:
+log%28+144%2F49+%29+=+2%2A%28+log%28+12+%29+-+log%28+7+%29+%29+
+log%28+144%2F49+%29+=+2%2A%28+.4771+%2B+.602+-+.845+%29+
( you can finish calculations )