How do you condense #logA - 2logB + 3logC#?

1 Answer
May 9, 2018

Answer:

#log((AC^3)/B^2)#

Explanation:

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx^nhArrnlogx#

#•color(white)(x)logx+logyhArrlog(xy)#

#•color(white)(x)logx-logyhArrlog(x/y)#

#rArrlogA-logB^2+logC^3#

#=log(A/B^2)+logC^3#

#=log((AC^3)/B^2)#