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Logx(8x-3)-logx 4=2
please help ill give you everything i have !!

asked
User Projeqht
by
7.8k points

1 Answer

1 vote
logx (8x-3) - logx 4 = 2
logx [(8x-3)/4] = 2
x^2=(8x-3)/4
4x^2=4(8x-3)/4
4x^2=8x-3
4x^2-8x+3=8x-3-8x+3
4x^2-8x+3=0
4(4x^2-8x+3=0)
(4^2)(x^2)-8(4x)+12=0
(4x)^2-8(4x)+12=0
(4x-2)(4x-6)=0
2(4x/2-2/2)2(4x/2-6/2)=0
4(2x-1)(2x-3)=0
4(2x-1)(2x-3)/4=0/4
(2x-1)(2x-3)=0
Two options:
1) 2x-1=0
2x-1+1=0+1
2x=1
2x/2=1/2
x=1/2

2) 2x-3=0
2x-3+3=0+3
2x=3
2x/2=3/2
x=3/2

Answer: Two solutions: x=1/2 and x=3/2
answered
User Guy Korland
by
7.3k points

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