How do you solve #3\log _ { 2} ( x - 1) + \log _ { 2} 4=5#?

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Answer 1 · 2018-01-10T09:16:41

#x=3#

Observe that in #3log_2(x-1)+log_2 4=5#, domain is #x>1#

#3log_2(x-1)+log_2 4=5# can be written as

#log_2 4(x-1)^3=5#

or #4(x-1)^3=2^5=32#

or #(x-1)^3=8# or #(x-1)^3-2^3=0#

as factors of #a^3-b^3# are #(a-b)(a^2+ab+b^2)#, we can write it as

#(x-1-2)((x-1)^2+2(x-1)+4)=0#

or #(x-3)(x^2-2x+1+2x-2+4)=0#

or #(x-3)(x^2+3)=0#

But as #x^2+3!=0#, wehave #x-3=0# or #x=3#