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

SOLUTION: Did I evaluate these log questions correctly? Or could I have simplified further? a) log5(4) + log5(2) =log5(42) =log5(8) b) 9 log3(9) + log3(81)^2 =9 log3(981)^2 =9lo

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) (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
=
=
=
=
= 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) (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

RELATED QUESTIONS
log5^3-log5^2-log5(1/4) as a single log attempt:... (answered by MathLover1,josmiceli)
log(x-4)+log5=2 (answered by Theo)
log5(x-2)= 2 + log5(x-4) (answered by lwsshak3)
log5(X-4)=... (answered by stanbon)
log5(x-1)=4+log5(x-2) (answered by nerdybill)
Solve:... (answered by lwsshak3)
log5(4)+log5(3) (answered by edjones)
log5(x-4)=-2 (answered by stanbon)
log5 x- log5... (answered by user_dude2008)