Did I solve this right? Should I factor?

Question

lo#g_2# (x+1) + lo#g_2# (x+7) = 3, =

lo#g_2# (x+1)(x+7)= 3, #2^3# = #x^2# + 8x + 7 =, 0 = #x^2# + 8x +7

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Answer 1 · 2017-11-28T07:53:14

No, it cannot be factorized. We should rather use quadratic formula. Solution is #x=0.14# or #7.14#.

Well it is #x^2+7x-1=0#, which cannot be factorized as discriminant is not a perfect square and we do not have rational roots.

#log_2(x+1)+log_2(x+7)=3# means domain of #x# is given by #x> -7#, as we cannot have logarithm of negative numbers.

#log_2(x+1)+log_2(x+7)=3# can be written as

#log_2(x+1)(x+7)=3#

or #(x+1)(x+7)=2^3=8#

or #x^2+8x+7=8#

or #x^2+7x-1=0#

or #x=(-7+-sqrt(7^2+4))/2=(-7+-sqrt53)/2#

or #x=(-7+sqrt53)/2=0.14# or #x=(-7-sqrt53)/2=7.14#

As both are within domain, solution is #x=0.14# or #7.14#