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Solve for x: ㏒(x+3)+㏒x=1

1 Answer

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Answer:

x=2

Explanation:

We can use the product rule of logarithms to simplify the equation:

㏒(x+3)+㏒x = 1

㏒[(x+3)x] = 1

(x+3)x = 10

Expanding the left side and rearranging the equation, we get:

x^2 + 3x - 10 = 0

We can now use the quadratic formula to solve for x:

x = (-b ± √(b^2 - 4ac)) / 2a

Where a = 1, b = 3, and c = -10

x = (-3 ± √(3^2 - 4(1)(-10))) / 2(1)

x = (-3 ± √49) / 2

x = (-3 ± 7) / 2

So x can be equal to -5 or 2. We need to check which solution, if any, is extraneous by plugging each one into the original equation and verifying that it is valid:

㏒(-5+3)+㏒(-5) ≠ 1

㏒(2+3)+㏒(2) = 1

Therefore, the solution to the equation is x = 2.

User Tmountjr
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