SOLUTION: Could someone please assist me by solving this logarithmic equation step-by-step for future reference. Thank you.
log(7y + 1) = 2log(y+3)-log2
log(7y + 1) = log(y+3)^2-log2
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-> SOLUTION: Could someone please assist me by solving this logarithmic equation step-by-step for future reference. Thank you.
log(7y + 1) = 2log(y+3)-log2
log(7y + 1) = log(y+3)^2-log2
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Question 104685: Could someone please assist me by solving this logarithmic equation step-by-step for future reference. Thank you.
log(7y + 1) = 2log(y+3)-log2
log(7y + 1) = log(y+3)^2-log2
log(7y + 1) = log(y+3)^2/2
you can write
(7y + 1) = (y+3)^2/2
2(7y + 1) = y^2+6y+9
14y+2=y^2+6y+9
y^2-8y+7=0
y1=(8+rootsquare(36))/2
=(8+6)/2
=7
y2=(8-rootsquare(36))/2
=(8-6)/2
=1
so (y1,y2) are (7,1) Found 2 solutions by scott8148, bimanewton:Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website! handy rules for logs:
adding logs is multiplying numbers
subtracting logs is dividing numbers
multiplying logs is exponentiation
dividing logs is taking roots
with those in mind ... 7y+1=((y+3)^2)/2 ... 14y+2=y^2+6y+9
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