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4 years ago

Solving Exponential and Logarithmic Equations In Exercise, solve for x. In 2x - In(3x - 1) = 0

Mathematics

1 answer:

LenKa [72]4 years ago

7 0

Answer:

$\frac{-1}{3x^2-x}$

Step-by-step explanation:

If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)

<u>using these information</u>:

g(x)=ln2x then g'(x)= $\frac{(2x)'}{2x} =\frac{2}{2x}=\frac{1}{x}$

h(x)=In(3x - 1) then h'(x)= $\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}$

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Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

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For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
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If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

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For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

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In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

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Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
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When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
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So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

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