How do you solve #log_11 2+2log_11x=log_11 32#?

Question

1131 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-27T20:43:15

#x=4#

#log_11 2 +2 log_11 x = log _ 11 32#

#log_11 2 +2 log_11 x - log _ 11 32=0#

#log_11 2 + log_11 x^2 - log _ 11 32=0#---> Use property #log_bx^n=nlog_bx#

#log_11 ((2x^2)/32)=0#---> Use properties #log_bx+log_by=log_b(xy)# and #log_bx-log_by=log_b(x/y)

#11^0=(x^2)/16#

#1=(x^2)/16#

#x^2=16#

#x=+- sqrt 16#

#x=+-4#

#:.x=4#