SOLUTION: Log5(3x^2-1)=Log5(2x)

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Question 176366This question is from textbook
: Log5(3x^2-1)=Log5(2x) This question is from textbook

Answer by EMStelley(208) About Me  (Show Source):
You can put this solution on YOUR website!
Since both of the sides are log base 5, all that is needed is to set the arguments equal to each other:
3x%5E2-1=2x
3x%5E2-2x-1
I will use the quadratic formula to solve with a=3,b=-2, c=-1.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%2A3%2A-1+%29%29%2F%282%2A3%29+
x+=+%282+%2B-+sqrt%284%2B12%29%29%2F6+=+%282+%2B-+sqrt%2816%29%29%2F6+=+%282+%2B-+4%29%2F6
So
x=6%2F6=1
and
x=-2%2F6=-1%2F3
Now, with logarithms you always want to check your answer because you cannot take the log of a negative number. Notice that if you plugged in x=-1/3 to the right hand side (log5(2x)) you would be taking the log of a negative number. So x=-1/3 is what is called an extraneous solution. Therefore, x=1 is the only solution.