How do you solve #log_2 x+log_8 32x=1#?

1 Answer
P dilip_k · ali ergin
Apr 23, 2016

#log_2x+log_8 32x=1#
#=>log_2x+log_8 32+log_8x=1#
#=>log_2x+log_8 2^5+log_8x=1#
#=>log_2x+5log_8 2+log_8x=1#
#=>log_2x+5xx1/log_2 8+1/log_x 8=1#
#=>log_2x+5xx1/log_2 2^3+1/log_x 2^3=1#
#=>log_2x+5xx1/(3log_2 2)+1/(3log_x 2)=1#
#=>log_2x+5/3+1/3xxlog_2 x=1#
#=>3log_2x+5+log_2 x=3#
#=>4log_2x=3-5=-2#
#=>log_2 x=-1/2#
#=>x=2^(-1/2)=1/sqrt2#

Related questions
See all questions in Logarithmic Models
Impact of this question
1675 views around the world
You can reuse this answer
Creative Commons License