SOLUTION: log(-x-3)-log(x-1)-log(x+3)= 1
<b>ALL LOGS ARE IN BASE 2!</b> (I couldn't find out how to put the base in the formula, sorry.)
I was thinking you would stack everything, like thi
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-> SOLUTION: log(-x-3)-log(x-1)-log(x+3)= 1
<b>ALL LOGS ARE IN BASE 2!</b> (I couldn't find out how to put the base in the formula, sorry.)
I was thinking you would stack everything, like thi
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Question 855108: log(-x-3)-log(x-1)-log(x+3)= 1 ALL LOGS ARE IN BASE 2! (I couldn't find out how to put the base in the formula, sorry.)
I was thinking you would stack everything, like this:
but I know that must not be right. Why do the two multiply at the bottom? It's supposed to be division/division/division (Log subtraction turns into division).
I'm more than a bit confused and appreciate all the help and explanation given. Thank you. Answer by josgarithmetic(39620) (Show Source):
How to put into base 2, and use the rendering tagging: log(-x-3)-log(x-1)-log(x+3)= 1;
log(2,(-x-3))-log(2,(x-1))-log(2,(x+3))=1.
Put three left-curly braces at the left and three right-curly braces at the right, and this will make the algebra dot com message field render the expression or equation.
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