Question 590236
1.	Show that log8x=1/3 log2x hence solve the equation for x>0
Log2(3x+1) +log8(x-1)3=0
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log8x=1/3 log2x
change to base2
log8x=log2x/log2(8)=log2x/3
..
log2(3x+1) +log8(x-1)*3=0
change to base 2
log2(3x+1) +log2(x-1)*3/log2(8)=0
log2(3x+1) +log2(x-1)*3/3=0
log2(3x+1) +log2(x-1)=0
place under single log
log2[(3x+1)(x-1)]=0
convert to exponential form:
2^0=(3x+1)(x-1)=1
3x^2-2x-2=0
solve by quadratic formula:
x≈-.549 (reject,x>0)
or
x=1.215