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What are the solutions to log (x2+8)= 1 +log (x)?

User Koja
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4 votes

Answer:

Explanation:

log(x²+8)=1+log(x)

log(x²+8)-log(x)=1


log(x^2+8)/(x) =1\\(x^2+8)/(x) =10^1\\x^2+8=10x\\x^2-10x+8=0\\x=(10 \pm √((-10)^2-4*1*8) )/(2) \\=(10 \pm √(100-32) )/(2) \\=(10 \pm √(68) )/(2) \\=(10 \pm 2√(17) )/(2) \\=5 \pm √(17)

User Zhong Huiwen
by
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