How do you solve #2\log x = \log 4+ \log ( x + 8)#?

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Answer 1 · 2017-07-28T12:10:15

Solution: # x =8 #

# 2 log x = log 4 + log ( x + 8) # or

# log x^2 = log (4* ( x + 8)) # or

# x^2 = 4x +32 or x^2 - 4x -32 =0 # or

# x^2 - 8x +4x -32 =0 # or

#x(x-8) + 4 (x-8) =0 # or

#(x-8) (x+ 4) =0 # Either #x-8=0 :. x =8 # or

#x+4=0 :. x =-4# . Check : for # x =-4 ; 2 log (-4) # is

invalid. Check : for # x =8 ; 2 log 8 = log 4 +log 16 # or

#log 64 = log 64 # (Valid )

So #x != -4#. Solution: # x =8 # [Ans]