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Lg(10x+200)=lg

{x}^(2)
+lg10​

1 Answer

4 votes

Answer: x = {-4, 5}

Explanation:

log (10x + 200) = log (x²) + log (10)

log (10x + 200) = log (10x²)

10x + 200 = 10x²

0 = 10x² - 10x - 200

0 = x² - x - 20

0 = (x - 5)(x + 4)

0 = x - 5 0 = x + 4

x = 5 x = -4

Check for extraneous solutions:

10x + 200 > 0 x² > 0

x = 5 10(5) + 200 > 0 (5)² > 0 both are satisfied

x = -4 10(-4) + 200 > 0 (-4)² > 0 both are satisfied

Both solutions are valid!

User Francesco Pitzalis
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