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wel
3 years ago
13

Solve: ln(x – 3) = ln(7x – 23) –  ln(x + 1)

Mathematics
1 answer:
Dvinal [7]3 years ago
8 0
Formula\ for\ lagarythm:\\
log_ab=c\ \ \ ----> a^c=b\\\\Base\ of\ ln \ is\ e=2,7\\\\ln(x-3)=ln(7x-23)-ln(x+1)\\\\
ln(x-3)=ln(\frac{7x-23}{x+1})\ \ \ | subtract\ ln(\frac{7x-23}{x+1})\\\\
ln(\frac{\frac{7x-23}{x+1}}{x-3})=0\\\\
ln(\frac{7x-23}{(x+1)(x-3)})=0\\\\
2,7^0=\frac{7x-23}{(x+1)(x-3)}\\\\
1=\frac{7x-23}{(x+1)(x-3)}\ \ \ | multiply\ by\ (x+1)(x-3)\\\\
(x+1)(x-3)=7x-23\\\\
x^2-3x+x-3=7x-23\\\\
x^2-9x-26=0\\\\x=\frac{1}{2}(9-\sqrt{185})\ \ or\ \ x=\frac{1}{2}(9+\sqrt{185})

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Answer:

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Step-by-step explanation:

We have been given a function f(x)=15x^3-15x^2-90x. We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

15x^3-15x^2-90x=0

Now, we will factor our equation. We can see that all terms of our equation a common factor that is 15x.

Upon factoring out 15x, we will get:

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Now, we will split the middle term of our equation into parts, whose sum is -1 and whose product is -6. We know such two numbers are -3\text{ and }2.

15x(x^2-3x+2x-6)=0

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Now, we will use zero product property to find the zeros of our given function.

15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0

15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0

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Step-by-step explanation:


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3 years ago
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