SOLUTION: Did I evaluate these log questions correctly? Or could I have simplified further? a) log5(4) + log5(2) =log5(4*2) =log5(8) b) 9 log3(9) + log3(81)^2 =9 log3(9*81)^2 =9lo

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Did I evaluate these log questions correctly? Or could I have simplified further? a) log5(4) + log5(2) =log5(4*2) =log5(8) b) 9 log3(9) + log3(81)^2 =9 log3(9*81)^2 =9lo      Log On


   

Question 973463: Did I evaluate these log questions correctly? Or could I have simplified further?
a) log5(4) + log5(2)
=log5(4*2)
=log5(8)
b) 9 log3(9) + log3(81)^2
=9 log3(9*81)^2
=9log3(729)^2
=9log3(531441)^2
Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a) log5(4) + log5(2)
=log5(4*2)
=log5(8)
You might add = log(8)/log(5)
-------------
b) 9 log3(9) + log3(81)^2
=====================
9 = 3^2, 81 = 3^4
= log%283%2C9%5E9%29+%2B+log%283%2C3%5E8%29
= log%283%2C3%5E18%29+%2B+log%283%2C3%5E8%29
= log%283%2C3%5E26%29
= 26%2Alog%283%2C3%29
= 26
=====================
=9 log3(9*81)^2
=9log3(729)^2
=9log3(531441)^2
----
531441 = 3^12
= 9log(3,3^24)
= log(3,3^26)
= 26

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
The first one is fine.
I'd probably simplify the second as I went along. Note: 729^2=531441, but no square. It has just been done.The first step you made has to have the 81^2 done before it is put in the parentheses. You would have 9 log3(9*81^2). It is easier to put the exponent in front so you have 9 log3(9) +2log3(81). The first is 18 (9*2) and the second is (4*2)=8



9log3 (9)=9*2=18
log3 (81^2)=2log 3(81) OR log3(6561)
log3(6561)=8
18+8=26