Question

please check!

Answer 1

@jdoe0001

Answer 2

well, recall that the number in front of it, is multiplying the whole expression it may expand to, so \(\bf log_4(3xyz)^2\implies 2log_4(3xyz)\implies 2[log_4(3xyz)]\\ \quad \\
2[log(3)+log(x)+log(y)+log(z)]\\ \quad \\2log(3)+2log(x)+2log(y)+2log(z)\)

Answer 3

hmm I missed the "4" in those fellows...lemme fix that \(\bf log_4(3xyz)^2\implies 2log_4(3xyz)\implies 2[log_4(3xyz)]\\ \quad \\
2[log_4(3)+log_4(x)+log_4(y)+log_4(z)]\\ \quad \\
2log_4(3)+2log_4(x)+2log_4(y)+2log_4(z)\)

Answer 4

but then when you multiplythe 2*2*2*2 you 32???

Answer 5

I mean 16*

Answer 6

the 2 in front of the expression as a single logarithm, it's just a coefficient
once the expression expands
is still a coefficient, but to the expanded version of the expression
thus it becomes a common factor to all terms in it

Answer 7

\(\bf log_5(5x^{-5})\implies log_5(5)+log_5(x^{-5})\implies log_5(5)+[-5log_5(x)]\\ \quad \\
log_5(5)-5log_5(x)\)

Answer 8

unless you meant \(\bf log_5(5x)^{-5}\)

Answer 9

No

Answer 10

ok