Question

How do you solve log(16+2b)=log(b2−4b)?

Answer 1

log(a)= log (b) ,a= b

16+ 2b = b^2 -4b

0=b^2-6b-16

0=(b-8)( b+2)

b=8 and -2

Answer 2

Solution;

log(16+2b)=log(b2−4b)

Remove log both sides, we get

=> 16+2b = b²-4b

=> 0= b²-4b-2b-16

=> 0 = b²-6b-16

=> b²-6b-16=0

Splitting the middle term,we get

=> b²-8b+2b-16=0

=> b(b-8)+2(b-8)=0

=> (b-8)(b+2) = 0

=> b-8=0 or b+2 = 0

=> b = 8 or b = -2

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