3 years ago

Solve for x, log(5x-5)=log(2x+3)+log(15)

1 answer:

5 0

$\log(5x-5)=\log(2x+3)+\log15\\
\log(5x-5)=\log[15(2x+3) ], \text{ because } \log(ab)=\log a+\log b\\
\log(5x-5)=\log(30x+45)\\
5x-5=30x+45, \text{ because } (\log x=\log y\rightarrow x=y.)\\
-25x=50\\
\text{Then } x=-2.$

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Help.... Other question got deleted.

Helen [10]

-2/25 is the slope
x1=4
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7 0

3 years ago

Find the complex zeros of x^3+27. write f in factored form

aliina [53]

$x^3+27=0
\\x^3+3^3=0
\\(x+3)(x^2-3x+9)=0
\\x^2-3x+9=0
\\a=1,b=-3,c=9
\\
\\x_{1,2}= \frac{-b\pm \sqrt{b^2-4ac} }{2a} = \frac{-(-3)\pm \sqrt{(-3)^2-4\times1\times9} }{2\times1} =\frac{3\pm \sqrt{-27} }{2} =\frac{3\pm 3\sqrt{3}i }{2} 
\\
\\(x+3)(x-\frac{3+ 3\sqrt{3}i }{2})(x-\frac{3- 3\sqrt{3}i }{2})=0

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