How do you solve #(log(x))^2=4#?

1 Answer
SagarStudy
Jun 14, 2016

#x=10^2# or #x=10^-2#

Explanation:

#(Log(x))^2=4#
#implies (Log(x))^2-2^2=0#

Use formula named as Difference of Squares which states that if #a^2-b^2=0#, then #(a-b)(a+b)=0#

Here #a^2=(Log(x))^2# and #b^2=2^2#

#implies (log(x)-2)(log(x)+2)=0#

Now, use Zero Product Property which states that if the product of two number, say #a# and #b#, is zero then one of two must be zero, i.e., either #a=0# or #b=0#.

Here #a=log(x)-2# and #b=log(x)+2#

#implies# either #log(x)-2=0# or #log(x)+2=0#
#implies# either #log(x)=2# or #log(x)=-2#
#implies# either #x=10^2# or #x=10^-2#

Related questions
See all questions in Common Logs
Impact of this question
11422 views around the world
You can reuse this answer
Creative Commons License