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3 years ago

Explain the process for solving this logarithmic equation.

1 answer:

7 0

Log3(x)+log3(4)-2log3(3)=2using properties of multiplication of logarithms,log3(x)+log3(4)-log3(3^2)=2log3(x)+log3(4)-log3(9)=2using properties of addition of logarithms,log3(x*4/9)=2raise both sides to exponent of base 33^log3(x*4/9)=3^24x/9=9x=81/4

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3 years ago

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What is the value of 4p -2 , when p=8 ?

Katen [24]

Answer:

Step-by-step explanation:

4(8)-2=32-2=30

Hope it helps! :)

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3 years ago

How are number lines and the coordinate plane related?

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3 years ago

What is the coefficient of -2x+8=10

RideAnS [48]

Step-by-step explanation:

Let's solve your equation step-by-step.

−2x+8=10

Step 1: Subtract 8 from both sides.

−2x+8−8=10−8

−2x=2

Step 2: Divide both sides by -2.

−2x−2=2−2

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Answer:

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3 years ago

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(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$