Express Logs In Terms of a, b, and c

[nycmathguy]

 

[MathLover1]

1. correction

log((100A^2)/(B^4*C^(1/3)))
(log((10^2)+log(A^2))-(log(B^4)+log(C^(1/3)))
(2log((10)+2log(A))-(4log(B)+(1/3)log(C))
2log((10)+2log(A)-4log(B)-(1/3)log(C)
2*1+2*a-4*b-(1/3)*c
2+2a-4b-(1/3)c
2 a - 4 b - c/3 + 2

2. correct
just simplify 5log(4)=5log(2^2)=5*2log(2) =10log(2)

 

[nycmathguy]

1. correction

log((100A^2)/(B^4*C^(1/3)))
(log((10^2)+log(A^2))-(log(B^4)+log(C^(1/3)))
(2log((10)+2log(A))-(4log(B)+(1/3)log(C))
2log((10)+2log(A)-4log(B)-(1/3)log(C)
2*1+2*a-4*b-(1/3)*c
2+2a-4b-(1/3)c
2 a - 4 b - c/3 + 2

2. correct
just simplify 5log(4)=5log(2^2)=5*2log(2) =10log(2)

It's hard not to make a mistake on the first problem.
In terms of simplifying 5log_10 (4) or 5 times log sub 10 of 4, are you sure this simplifies to 10log(2)?

 

[MathLover1]

yes, as I stated above
5log_10 (4) =5log_10 (^2) =5*2log_10 (2) =10_10log(2)

 

[nycmathguy]

yes, as I stated above
5log_10 (4) =5log_10 (^2) =5*2log_10 (2) =10_10log(2)

Ok. Thanks. A little precalculus review. Believe it or not, I miss precalculus.