Solve for x using properties of logarithms: log2(32)-log3(x)=log5(25)?

2 Answers
Alan P.
Feb 22, 2018

#x=1#

Explanation:

Since #2^5=3color(white)("xx")harrcolor(white)("xx")log_2(32)=5#
and #color(white)("x")5^2=25color(white)("xx")harrcolor(white)("xx")log_5(25)=2#

#log_2(32)-log_3(x)=log_5(25)#
is equivalent to
#5-log_3(x)=2#

#rArr log_3(x)=3#
and
since #3^color(blue)1=3#
#color(white)("XXX")log_3(x)=3color(white)("xx")rarrcolor(white)("xx")x=color(blue)1#

Cem Sentin
Feb 22, 2018

#x=27#

Explanation:

#log_2(32)-log_3(x)=log_5(25)#

#log_2(2^5)-log_3(x)=log_5(5^2)#

#5log_2(2)-log_3(x)=2log_5(5)#

#5-log_3(x)=2#

#log_3(x)=3#

Hence #x=3^3=27#

Related questions
Impact of this question
2198 views around the world
You can reuse this answer
Creative Commons License