# Given #log_3(5x^2+x−y)=1# then #y=ax^n+x+b#, what is #a#, #b#, #n#?

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See explanatio.

So we can write that:

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Given ( \log_3(5x^2 + x - y) = 1 ), we can rewrite the equation in exponential form to solve for (y):

[ 3^1 = 5x^2 + x - y ]

This simplifies to:

[ y = 5x^2 + x - 3 ]

Comparing this to the general form ( y = ax^n + x + b ), we find:

- ( a = 5 )
- ( n = 2 )
- ( b = -3 )

Therefore, (a = 5), (b = -3), and (n = 2).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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