Log_4*log_3*log_2(x^2-2x)=0 X=?

1 Answer
Jun 27, 2018

#x=0#, #x=2#

Explanation:

Factor log first as shown below:

#log(4).log(3).log2(x^2-2x)=0#

Take #x^2-2x# = #x(x-2)#

So now we get:

#log(4).log(3).log(2).x(x-2)=0#

Simplify #log4# = #log2^2# = #2log2#

Remember log rule: #log_a(x)^b# = #blog_x(x)#

So now we get:

#2log(2).log(3).log(2).x(x-2)=0#

Apply exponential rule:

#a^b.a^c# = #a^(b+c)#

#log(2) .log(2)# = #log2^(1+1)# = #log2^2#

And we get:

#2log(2).log(2)^2.log(3)x(x-2)=0#

Using the zero factor principle, #x=0#

Solve #x-2=0#
we get #x=2#

So value of #x=0#, #x=2#

Hope this helps!