Question 563646
log2(x+1)=log4(7x-5)
convert to log base 2
log2(x+1)=log2(7x-5)/log2(4)
log2(x+1)=log2(7x-5)/2
2log2(x+1)-log2(7x-5)=0
log2(x+1)^2-log2(7x-5)=0
place under single log
log2[(x+1)^2/(7x-5)]=0
convert to exponential form: Base(2) raised to log of number(0)=number[(x+1)^2/(7x-5)]
2^0=[(x+1)^2/(7x-5)]=1
[(x+1)^2=(7x-5)]
x^2+2x+1=7x-5
x^2-5x+6=0
(x-2)(x-3)=0
x=2
or
x=3