How do you solve #ln(x - 3) + ln(x + 4) = 1#?
Now, that we have the domain we can solve the equation in this domain, thus:
We apply the quadratic formula,
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To solve the equation ln(x - 3) + ln(x + 4) = 1, you can use the properties of logarithms to combine the two logarithmic terms into a single logarithm. Then, you can exponentiate both sides to eliminate the logarithm. After that, you can solve for x. Here are the steps:
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Combine the logarithmic terms using the property: ln(a) + ln(b) = ln(a * b). ln(x - 3) + ln(x + 4) = ln((x - 3)(x + 4)).
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Rewrite the equation: ln((x - 3)(x + 4)) = 1.
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Exponentiate both sides using the property of logarithms: e^ln(u) = u. e^ln((x - 3)(x + 4)) = e^1.
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Simplify: (x - 3)(x + 4) = e.
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Expand and rearrange the equation: x^2 + x - 12 - e = 0.
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Solve the quadratic equation for x using the quadratic formula or factoring.
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Once you find the solutions for x, check if they are valid by ensuring they satisfy the domain of the original logarithmic equation, which requires x > 3 and x > -4.
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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.
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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