How do you solve #log_b9+log_bx^2=log_bx#?
that's equivalent to:
and the solution is:
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To solve the equation (\log_b 9 + \log_b x^2 = \log_b x), you can use the properties of logarithms.

Apply the product rule of logarithms to combine the logarithms on the left side: (\log_b 9x^2 = \log_b x).

Since the bases are the same, the expressions inside the logarithms must be equal: (9x^2 = x).

Rearrange the equation to form a quadratic equation: (9x^2  x = 0).

Factor out the common factor, (x): (x(9x  1) = 0).

Set each factor equal to zero and solve for (x): (x = 0) or (9x  1 = 0).

Solve (9x  1 = 0) for (x): (9x = 1) (\Rightarrow) (x = \frac{1}{9}).
Therefore, the solutions are (x = 0) and (x = \frac{1}{9}).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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