Log8+log2
How Do You Solve Log1+ Log2+ Log3+ Log4+ Log5+ Log6+ Log7+ Log8+ Log9+ Log10 By Hand, When Given Log 2=0.301,log3=0.4777,and Log 7=0.845?
Log[1] = 0
Log[2] is given
Log[3] is given
Log[4] = Log[2^2] = 2*Log[2]
Log[5] = Log[10/2] = Log[10] - Log[2] = 1 - Log[2]
Log[6] = Log[2*3] = Log[2] + Log[3]
Log[7] is given
Log[8] = Log[2^3] = 3*Log[2]
Log[9] = Log[3^2] = 2*Log[3]
Log[10] = 1
Your sum is
0 + Log[2] + Log[3] + 2*Log[2] + (1 - Log[2]) + (Log[2] + Log[3]) + Log[7] + 3*Log[2] + 2*Log[3] + 1
= (1+2-1+1+3)*Log[2] + (1+1+2)*Log[3] + Log[7] + 2
= 6*Log[2] + 4*Log[3] + Log[7] + 2
Log[2] is given
Log[3] is given
Log[4] = Log[2^2] = 2*Log[2]
Log[5] = Log[10/2] = Log[10] - Log[2] = 1 - Log[2]
Log[6] = Log[2*3] = Log[2] + Log[3]
Log[7] is given
Log[8] = Log[2^3] = 3*Log[2]
Log[9] = Log[3^2] = 2*Log[3]
Log[10] = 1
Your sum is
0 + Log[2] + Log[3] + 2*Log[2] + (1 - Log[2]) + (Log[2] + Log[3]) + Log[7] + 3*Log[2] + 2*Log[3] + 1
= (1+2-1+1+3)*Log[2] + (1+1+2)*Log[3] + Log[7] + 2
= 6*Log[2] + 4*Log[3] + Log[7] + 2
