Please help me with this logarithmic equation. How do I solve it?

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Answer 1 · 2018-05-28T12:52:02

Solution : # x=4#

# log_8 (x+4) + log_8 (x-3) = 1 # or

# log_8 {(x+4)(x-3)} =log_8 8 [log_b b=1]# or

# (x+4)(x-3) = 8 or x^2 +x -12 -8 =0# or

#x^2 +x -20 =0 or (x+5)(x-4)=0# or

#x= -5 , x =4# When #x= -5 ; x+4= -1# (negative)

#:.x=-5# is invalid . Solution is # x=4# [Ans]

Answer 2 · 2018-05-28T12:52:35

#x=4#

Writing
#log_8(x+4)(x-3)=log_8 8#
so
#(x+4)(x-3)=8#
#x^2+4x-3x-12=8#
thus
#x^2+x-20=0#
with the Solutions
#x_1=4#
#x_2=-5# (which is not a solution!)