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Check for extraneous lnx+ln(x+2)=ln(x+6)

User Hsuk
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Final answer:

To check for extraneous solutions in the equation lnx + ln(x+2) = ln(x+6), we substitute the possible solution back into the equation. By simplifying the equation and solving for x, we find that x = 2 is the valid solution.

Step-by-step explanation:

To check for extraneous solutions, we need to substitute the possible solution back into the original equation and see if we get a valid solution. In this case, we need to substitute the value of x into the equation: lnx + ln(x+2) = ln(x+6).

We can simplify the equation by using the properties of logarithms. Adding two logarithms is the same as multiplying the arguments: ln(x(x+2)) = ln(x+6).

Now we can equate the arguments of the logarithms: x(x+2) = x+6.

Simplifying the equation further, we get a quadratic equation: x^2 + 2x = x + 6.

After rearranging the terms and simplifying, we have: x^2 + x - 6 = 0. Factorizing or using the quadratic formula, we find the solutions: x = -3 or x = 2.

To check if these solutions are extraneous, we substitute them back into the original equation. Plugging in x = -3 gives ln(-3) + ln(-3+2) = ln(-3+6), which is not defined since the natural logarithm is only defined for positive numbers. Therefore, x = -3 is an extraneous solution. Plugging in x = 2 gives lnn(2) + ln(2+2) = ln(2+6), which is true. Thus, the solution to the equation is x = 2.

User Sqwerty
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