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Julli [10]
3 years ago
10

Find X : log2+ logx=1

Mathematics
2 answers:
Dominik [7]3 years ago
8 0
log2+logx=1;\ D:x\in\mathbb{R^+}\\\\log(2\ \cdot\ x)=log10\iff2x=10\ \ \ /:2\\\\x=5\in D\\\\Solution:x=5.
Vinil7 [7]3 years ago
3 0
\log2+\log x=1\\
D:x>0\\
\log2x=1\\
10^1=2x\\
2x=10\\
x=5
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Which equation has the solutions X=1 +-SQRT 5?
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Answer:

D is correct. x^2-2x-4=0

Step-by-step explanation:

We are given the root of the equation

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If a and b are the solution of equation then equation would be (x-a)(x-b)=0

Here, a=1+\sqrt{5} , b=1-\sqrt{5}

Equation form would be (x-1-\sqrt{5})(x-1+\sqrt{5})=0

Now we simplify the above equation to get correct option.

x^2-x+x\sqrt{5}-x+1-\sqrt{5}-x\sqrt{5}+\sqrt{5}-5=0

x^2-2x-4=0

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4 years ago
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emmasim [6.3K]
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5 weeks

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$16120000 was achieved in 52 weeks

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