Solve-2-x-x-4-

Tinku Tara June 4, 2023 Algebra 0 Comments

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Question Number 15570 by tawa tawa last updated on 11/Jun/17

Solve: 2^x = x^4

Solve : 2^{x} = x^{4}

Commented by tawa tawa last updated on 11/Jun/17

workings

workings

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17

Commented by mrW1 last updated on 12/Jun/17

the graph is much bigger than what you display. there is a third zero point on the right side.

the graph is much bigger than what

you display . there is a third zero

point on the right side .

Commented by tawa tawa last updated on 12/Jun/17

God bless you sir.

God bless you sir .

Answered by mrW1 last updated on 12/Jun/17

2^x =x^4 2^(x/4) =±x e^((x/4)ln 2) =±x ±1=xe^(−((xln 2)/4)) ∓((ln 2)/4)=(−((xln 2)/4))e^((−((xln 2)/4))) ⇒−((xln 2)/4)=W(∓((ln 2)/4)) x=((W(∓((ln 2)/4)))/(−((ln 2)/4)))=((W(∓0.173287))/(−0.173287)) = { ((((−2.772589)/(−0.173287))=16)),((((−0.214811)/(−0.173287))=1.2396)),((((0.149260)/(−0.173287))=−0.8613)) :}

2^{x} = x^{4}

2^{\frac{x}{4}} = \pm x

e^{\frac{x}{4} \ln 2} = \pm x

\pm 1 = {xe}^{- \frac{xln 2}{4}}

\mp \frac{\ln 2}{4} = (- \frac{xln 2}{4}) e^{(- \frac{xln 2}{4})}

\Rightarrow - \frac{xln 2}{4} = W (\mp \frac{\ln 2}{4})

x = \frac{W (\mp \frac{\ln 2}{4})}{- \frac{\ln 2}{4}} = \frac{W (\mp 0.173287)}{- 0.173287}

= {\begin{cases} \frac{- 2.772589}{- 0.173287} = 16 \\ \frac{- 0.214811}{- 0.173287} = 1.2396 \\ \frac{0.149260}{- 0.173287} = - 0.8613 \end{cases}

Commented by tawa tawa last updated on 12/Jun/17

God bless you sir.

God bless you sir .

Commented by Tinkutara last updated on 12/Jun/17

Sir, how do you calculate W(± 0.173287)? Is there any method to calculate this without calculator? How to find this using Geogebra?

Sir, how do you calculate W (\pm 0.173287) ?

Is there any method to calculate this

without calculator ? How to find this

using Geogebra ?

Commented by mrW1 last updated on 12/Jun/17

Normal calculators don′t have this Lambert function. There are some online calculators for W−function, but most of them give only one value for x<0. With geogebra you can calculate W−function value by yourself: For example you want to calculate W(−((ln 2)/4)). That is the root(s) of f(x)=xe^x +((ln 2)/4).

Normal calculators {don}^{'} t have this

Lambert function . There are some

online calculators for W - function,

but most of them give only one value

for x < 0 .

With geogebra you can calculate

W - function value by yourself :

For example you want to calculate

W (- \frac{\ln 2}{4}) . That is the root (s) of

f (x) = {xe}^{x} + \frac{\ln 2}{4} .

Commented by mrW1 last updated on 12/Jun/17

Commented by Tinkutara last updated on 12/Jun/17

Do you mean that in general if we want W(n), these are the roots of f(x) = xe^x − n ?

Do you mean that in general if we

want W (n), these are the roots of

f (x) = {x e}^{x} - n ?

Commented by mrW1 last updated on 12/Jun/17

yes. this is the definition of W function.

yes . this is the definition of W function .

Commented by Tinkutara last updated on 12/Jun/17

Thanks Sir!

Thanks Sir!

Commented by tawa tawa last updated on 12/Jun/17

logx^6 = (x/6) 6logx = x/6 36logx = x x^(36) = e^x 1 = x^(36) e^(−x) log .... Sir, i have been thinking. i have a question. What is my aim in solving using lambert function. Please give example as the number 1, i will use it to solve number 2. my problem here is that i dont even have starting point. Thank you sir. I want to learn it. The starting point is my problem. please do number 1 as example.

{logx}^{6} = \frac{x}{6}

6 logx = x / 6

36 logx = x

x^{36} = e^{x}

1 = x^{36} e^{- x}

\log \dots .

Sir, i have been thinking .

i have a question .

What is my aim in solving using lambert function .

Please give example as the number 1, i will use it to solve number 2 .

my problem here is that i dont even have starting point .

Thank you sir . I want to learn it .

The starting point is my problem .

please do number 1 as example .

Commented by tawa tawa last updated on 12/Jun/17

Please give me one question on this, let me solve it.

Please give me one question on this, let me solve it .

Commented by mrW1 last updated on 12/Jun/17

Solve x (1) log (x^6 )=(x/6) (2) log (x^3 )=(x/3)

Solve x

(1) \log (x^{6}) = \frac{x}{6}

(2) \log (x^{3}) = \frac{x}{3}

Commented by mrW1 last updated on 12/Jun/17

To be able to use Lambert function we must re−arange the original equation into following form: A×e^A =B wherein A is a expression with x and B is a expression with constants only. The solution is then A=W(B) Example 1: log x^6 =(x/6) ⇒6log (±x)=(x/6) (± because x can be + or −) we consider +x at first ⇒log (x)=(x/(36)) ⇒((ln x)/(ln 10))=(x/(36)) ⇒ln x=((ln 10)/(36))x ⇒x=e^((ln 10 x)/(36)) ⇒xe^(−((ln 10 x)/(36))) =1 ⇒(−((ln 10 x)/(36)))e^(−((ln 10 x)/(36)) ) =−((ln 10)/(36)) this is the form A×e^A =B we need. the solution is −((ln 10 x)/(36))=W(−((ln 10)/(36))) ⇒x=−((36)/(ln 10))×W(−((ln 10)/(36))) we will get here 2 values for x, since W(−((ln 10)/(36))) has 2 values. similarly for the case −x we have ⇒log (−x)=(x/(36)) ⇒((ln (−x))/(ln 10))=(x/(36)) ⇒ln (−x)=((ln 10)/(36))x ⇒−x=e^((ln 10 x)/(36)) ⇒−xe^(−((ln 10 x)/(36))) =1 ⇒(−((ln 10 x)/(36)))e^(−((ln 10 x)/(36)) ) =((ln 10)/(36)) ⇒−((ln 10 x)/(36))=W(((ln 10)/(36))) ⇒x=−((36)/(ln 10))×W(((ln 10)/(36))) we will get here only one value for x. All solutions are thus x= { ((−((36)/(ln 10))×W(−((ln 10)/(36))) → 2 values)),((−((36)/(ln 10))×W(((ln 10)/(36))) → 1 value)) :}

To be able to use Lambert function

we must re - arange the original

equation into following form :

A \times e^{A} = B

wherein A is a expression with x and

B is a expression with constants only .

The solution is then

A = W (B)

Example 1 : \log x^{6} = \frac{x}{6}

\Rightarrow 6 \log (\pm x) = \frac{x}{6} (\pm because x can be + or -)

we consider + x at first

\Rightarrow \log (x) = \frac{x}{36}

\Rightarrow \frac{\ln x}{\ln 10} = \frac{x}{36}

\Rightarrow \ln x = \frac{\ln 10}{36} x

\Rightarrow x = e^{\frac{\ln 10 x}{36}}

\Rightarrow {xe}^{- \frac{\ln 10 x}{36}} = 1

\Rightarrow (- \frac{\ln 10 x}{36}) e^{- \frac{\ln 10 x}{36}} = - \frac{\ln 10}{36}

this is the form A \times e^{A} = B we need .

the solution is

- \frac{\ln 10 x}{36} = W (- \frac{\ln 10}{36})

\Rightarrow x = - \frac{36}{\ln 10} \times W (- \frac{\ln 10}{36})

we will get here 2 values for x, since

W (- \frac{\ln 10}{36}) has 2 values .

similarly for the case - x we have

\Rightarrow \log (- x) = \frac{x}{36}

\Rightarrow \frac{\ln (- x)}{\ln 10} = \frac{x}{36}

\Rightarrow \ln (- x) = \frac{\ln 10}{36} x

\Rightarrow - x = e^{\frac{\ln 10 x}{36}}

\Rightarrow - {xe}^{- \frac{\ln 10 x}{36}} = 1

\Rightarrow (- \frac{\ln 10 x}{36}) e^{- \frac{\ln 10 x}{36}} = \frac{\ln 10}{36}

\Rightarrow - \frac{\ln 10 x}{36} = W (\frac{\ln 10}{36})

\Rightarrow x = - \frac{36}{\ln 10} \times W (\frac{\ln 10}{36})

we will get here only one value for x .

All solutions are thus

x = {\begin{cases} - \frac{36}{\ln 10} \times W (- \frac{\ln 10}{36}) \to 2 values \\ - \frac{36}{\ln 10} \times W (\frac{\ln 10}{36}) \to 1 value \end{cases}

Commented by Tinkutara last updated on 12/Jun/17

In the last line how can we determine that W(((−ln 10)/(36))) has 2 values while W(((ln 10)/(36))) has one value? Can it be without Geogebra?

In the last line how can we determine

that W (\frac{- \ln 10}{36}) has 2 values while

W (\frac{\ln 10}{36}) has one value ? Can it be

without Geogebra ?

Commented by mrW1 last updated on 12/Jun/17

W(x) is the inverse function of f(x)=xe^x W(x) has only one value if x≥0 and 2 values if −(1/e)<x<0.

W (x) is the inverse function of f (x) = {xe}^{x}

W (x) has only one value if x ⩾ 0

and 2 values if - \frac{1}{e} < x < 0 .

Commented by mrW1 last updated on 12/Jun/17

Commented by tawa tawa last updated on 12/Jun/17

2) ln(x^3 ) = (x/3) 3ln(±x) = (x/3) ln(±x) = (x/9) ((ln(±x))/(ln(10))) = (x/9) ln(±x) = ((xln(10))/9) ± x = e^((xln(10))/9) 1 = ±xe^(−((xln(10))/9)) −((xln(10))/9)e^(−((xln(10))/9)) = −((ln10)/9) −((xln(10))/9) = W[−((ln(10))/9)] x = − (9/(ln(10))) W[−((ln(10))/9)] x = − ((9(−0.3706))/(ln(10))) or x = ((9(−2.110))/(ln(10)))

2)

\ln (x^{3}) = \frac{x}{3}

3 \ln (\pm x) = \frac{x}{3}

\ln (\pm x) = \frac{x}{9}

\frac{\ln (\pm x)}{\ln (10)} = \frac{x}{9}

\ln (\pm x) = \frac{xln (10)}{9}

\pm x = e^{\frac{xln (10)}{9}}

1 = \pm {xe}^{- \frac{xln (10)}{9}}

- \frac{xln (10)}{9} e^{- \frac{xln (10)}{9}} = - \frac{\ln 10}{9}

- \frac{xln (10)}{9} = W [- \frac{\ln (10)}{9}]

x = - \frac{9}{\ln (10)} W [- \frac{\ln (10)}{9}]

x = - \frac{9 (- 0.3706)}{\ln (10)} or x = \frac{9 (- 2.110)}{\ln (10)}

Commented by tawa tawa last updated on 12/Jun/17

God bless you sir.

God bless you sir .

Commented by mrW1 last updated on 12/Jun/17

in question (1) in log x^6 x can be + or −, thus log x^6 =6log (±x) we have totally 3 solutions. but in question (2) in log x^3 x must be +, thus log x^3 =3log x we have totally 2 solutions: x=−(9/(ln 10))×W(−((ln 10)/9))

in question (1) in \log x^{6} x can be + or -,

thus \log x^{6} = 6 \log (\pm x)

we have totally 3 solutions .

but in question (2) in \log x^{3} x must be +,

thus \log x^{3} = 3 \log x

we have totally 2 solutions :

x = - \frac{9}{\ln 10} \times W (- \frac{\ln 10}{9})

Commented by tawa tawa last updated on 12/Jun/17

Thank you sir. i understand.

Thank you sir . i understand .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17

thank you so much dear mrW1. it is very beautiful and useful. god blees you master.

t h a n k y o u s o m u c h d e a r m r W 1 .

i t i s v e r y b e a u t i f u l a n d u s e f u l .

g o d b l e e s y o u m a s t e r .

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