How do you solve #\log _ { 4} ( 9v + 2) = \log _ { 4} ( v ^ { 2} + 2)#?
1 Answer
Jan 12, 2018
Explanation:
#"using the "color(blue)"law of logarithms"#
#•color(white)(x)log_bx=log_byrArrx=y#
#rArrv^2+2=9v+2#
#rArrv^2-9v=0larrcolor(blue)"in standard form"#
#rArrv(v-9)=0#
#rArrv=0" or "v=9#
