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3 years ago

Solving Exponential and Logarithmic Equations In Exercise, solve for x. 500(1.075)120x = 100.000

Mathematics

1 answer:

Volgvan3 years ago

4 0

Answer:

The solution is:

$x = 0.61$

Step-by-step explanation:

The first step to solve this equation is placing everything with the exponential to one side of the equality, and everything without the exponential to the other side. So

$500(1.075)^{120x} = 100000$

$(1.075)^{120x} = \frac{100000}{500}$

$(1.075)^{120x} = 200$

To find x, we have to apply log to both sides of the equality.

We also have that:

$\log{a^{x}} = x\log{a}$

$\log{(1.075)^{120x}} = \log{200}$

$120x\log{1.075} = 2.30$

$120x*0.03 = 2.30$

$3.77x = 2.30$

$x = \frac{2.30}{3.77}$

$x = 0.61$

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Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

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Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

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Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$

so the new equation is

$y''(t)+ 8y'(t)-20y(t)=0$

the auxiliary equation is

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so the solutions of the new equation are

$y(t)=ae^{2t}+be^{-10t}$

with a and b real

$x(t)=e^t\\ t(x)=ln(x)$

$y(x)=ae^{2ln(x)}+be^{-10ln(x)}=ax^2+bx^{-10}$

hope this helps

do not hesitate if you have any questions

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