SOLUTION: 30) 7log7(x+3) I've no idea what the 7 means before the log7 31) log2 x = 5 Is this one really just 2^5 = X? Then solve for x = 32?

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: 30) 7log7(x+3) I've no idea what the 7 means before the log7 31) log2 x = 5 Is this one really just 2^5 = X? Then solve for x = 32?       Log On


   

Question 586321: 30) 7log7(x+3)
I've no idea what the 7 means before the log7
31) log2 x = 5
Is this one really just 2^5 = X?
Then solve for x = 32?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
30) 7log7(x+3)
I've no idea what the 7 means before the log7
I think the 7 before the log 7 can be just a multiplier or an exponent for the number (x+3)
The equation can be rewritten as: log7(x+3)^7
..
31) log2 x = 5
Is this one really just 2^5 = X?
Then solve for x = 32?
You got this one right. In plain English it reads: Base(2) raised to the log of the number(5) = number.