Question 458934: log(base2)8 + 2log(base2)x = log(base2)(14x-3)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! log(base2)8 + 2log(base2)x = log(base2)(14x-3)
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log2(8)+2log2(x)=log2(14x-3)
log2(8)+2log2(x)-log2(14x-3)=0
log2(8x^2/14x-3)=0
convert to exponential form: (base(2) raised to log of number(0)=number(8x^2/14x-3)
2^0=(8x^2/14x-3)=1
8x^2=14x-3
8x^2-14x+3=0
solve by quadratic formula
a=8, b=-14, c=3
x=[-(-14)ħsqrt(-14)^2-4*8*3]2*8
x=[14ħsqrt(196-96]/16
x=(14ħ√100)/16=14ħ10/16
x=24/16=1.5
or
x=4/16=.25
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