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3 years ago

How do you solve x+4=10^x

Mathematics

1 answer:

raketka [301]3 years ago

3 0

Two ways to solve the problem:

A. By graphing f(x)=10^x and g(x)=x+4.
The intersection(s) will be the solution.
See graph below, approximate solutions: (-4,0),(0.66, 4.67)

B. Refine approximate solutions using Newton's method
Let h(x)=f(x)-g(x) = 10^x-x-4 ...............(1)
we calculate the derivative, h'(x) = log(10)*10^x-1 [ note:log(x) means ln(x) ]
and use Newton's iterative formula to find successive approximations to the root, basically refining the approximate solutions.
The iterative formula for nth approximation $x_n$ is given by
$x_n$ = $x_{n-1}$ - h( $x_n$ ) / h'( $x_n$ )....(2)

Using initial approximation (-4,0), we have x0=-4
x1
=x0-h(x0)/h'(x0)
=-4 - h(-4)/h'(-4)
=-4 - (10^(-4)-(-4)-4)/(log(10*10^(-4)-1)
=-4 - (1/10000)/(log(10)/10000-1)
=-3.9999000

Repeating the same for second approximation, x2, we get
x2=-3.99989997696619 which is accurate to 14 places after decimal

Now we can refine the other approximate solution, x0=0.66 to find
x1=0.669356
x2=0.6692468481102326
x3=0.669246832877748
x4=0.6692468328777476
So we will accept x=0.669246832877748

So the solutions are S={-3.99989997696619,0.6692468328777476} both accurate to 14 places after the decimal

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3 years ago

Solve the equation, and check the solutions (√x+3)+6=3.

larisa86 [58]

Answer:

The equation has no solution.

Step-by-step explanation:

Hi there!

First, let´s write the equation:

(√x + 3) + 6 = 3

To solve the equation, let´s subtract 6 to both sides of the equation:

√x + 3 = -3

subtract 3 to both sides of the equation:

√x = -6 (the square root of any number can´t be negative, but let´s solve it anyway and then check the solution).

Square both sides of the equation:

x = 36

Now, let´s check the solution:

(√x + 3) + 6 = 3

x = 36

(√36 + 3) + 6 = 3

(6 + 3) + 6 = 3

9 + 6 = 3

15 ≠ 3

Then, the equation has no solution.

Have a nice day!

6 0

3 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.