Humphrey77: @ smurfy please expand 2((2X-3)(5X+1)(X+7)(X+2))

I got my original question. I told you I didn't see the 20. I thought it was 2. Sorry about that

hardedeji1: can anyone help with the solution to this?

3^x + 4^x = 5^x

OBIVIOUSLY 4-1 -is 3 .

we know that: (4-1)raise to power x + (4)raise to power x equals (4+1 ) raise to power x
clearly WE HAVE( 4(1-1/4))raise to power x + (4)raise to power x equal to (4(1+1/4) raise to power x . by linear approximation (1+x) raise to power n we have 1+


nx . when x is very small ( less than 1) clearly we have

4raise to power x( 1-x/4) + 4raise to power x equal to 4 raise to power x(1+x/4) so clearly we obtain the algebraic equation : 1-x/4 +1 is equal to 1+x/4 clearly we obtain
2-x/4 is equal to 1 + x/4
collect like term we have x equal to 2

smurfy: Factorise completely 20x^4 + 154x^3 + 40x^2 - 418x - 84.

the aswer is
2(x+2)(x+7)(2x-3)(5x+1). shikena.

maths guru please what is the solution to the equation 6"x-31(2"x)+32 equal to zero . (.please note the symbol " means raise to power)

benbuks: If (logx _(4))^2 =(logx_(2) )(log a_(x) , find a

...{logx(base4)}^2=logx(base2)*logx(basea)...{logx(base4)}^2=logx(base4)*logx(base4)...then.logx(base4)*logx(base4)=logx(base2)*logx(basea).....logx(basea)=logx(base4)*logx(base4)/logx(base2)...(i)...{logx(base4)/logx(base2)=1/2}...logx(basea)=1/2.logx(base4)....logx(basea)=logx(base16)...comparing lhs and rhs...a=16

maths guru please what is the solution to the equation 6"x-31(2"x)+32 equal to zero . (.please note the symbol " means raise to power) ie 6 raise to power x -31(2 raise to power x) +32 equal to zero

Calculusf(x):
...{logx(base4)}^2=logx(base2)*logx(basea)...{logx(base4)}^2=logx(base4)*logx(base4)...then.logx(base4)*logx(base4)=logx(base2)*logx(basea).....logx(basea)=logx(base4)*logx(base4)/logx(base2)...(i)...{logx(base4)/logx(base2)=1/2}...logx(basea)=1/2.logx(base4)....logx(basea)=logx(base16)...comparing lhs and rhs...a=16

GOOD ! IT CAN ALSO BE OF THE FORM a raise to power log x (base 16) which is equalvent as x . so clearly a is 16

smurfy:

Here goes...

Y = x^x^x
y = (x^x)^x
Let u = x^x
Take log of both sides:
log(u) = log(x^x)
log(u) = xlogx
d/dx[log(u)] = d/dx(xlogx)
1/u(du/dx) = logx + 1
du/dx = u(logx + 1)
du/dx = (x^x)(logx + 1)

Recall that y = u^x
y = u^x
logy = log(u^x)
logy = xlog(u)
d/du(logy) = d/du[xlog(u)]
dy/du(1/y) = x/u
dy/du = xy/u
dy/du = x(x^x^x)/(x^x)

Now, dy/dx = dy/du * du/dx

dy/dx = x(x^x^x)/(x^x) * (x^x)(logx + 1)

dy/dx = x(x^x^x) * (logx + 1)

.i respect you bro,but here is my approach...y=x^x^x...y=x^(x^2).try this with calculator and substitute values like 2 and 3 and check if it's correct...for 2...x^x^x=16 and x^(x^2)=16 try that of 3 also...and y=x^x^x means...y={[(x)^x]^x} and from indices u=(x^a)^b=x^ab...then y=x^(x.x)=x^(x^2)...so from the question y=x^x^x...taken natural log of both sides...lny=ln.x^x^x...lny=xlnx^x...(lnx^x=xlnx)then lny=x.xlnx...lny=x^2lnx...1/y.dy/dx=x+2xlnx...dy/dx=x^x^x{x+2xlnx}

Humphrey77: GOOD ! IT CAN ALSO BE OF THE FORM a raise to power log x (base 16) which is equalvent as x . so clearly a is 16

.i respect you bro...particularly that ur real analysis...you are great....and concerning ur question,instead of re-typing it,you supposed to modify it...

Calculusf(x):
...{logx(base4)}^2=logx(base2)*logx(basea)...{logx(base4)}^2=logx(base4)*logx(base4)...then.logx(base4)*logx(base4)=logx(base2)*logx(basea).....logx(basea)=logx(base4)*logx(base4)/logx(base2)...(i)...{logx(base4)/logx(base2)=1/2}...logx(basea)=1/2.logx(base4)....logx(basea)=logx(base16)...comparing lhs and rhs...a=16

...great Fx..i knew you will kill it..

find the missing numbers : 2,_10,_40 ( hint using commutative law of addition)

Calculusf(x):
.i respect you bro,but here is my approach...y=x^x^x...y=x^(x^2).try this with calculator and substitute values like 2 and 3 and check if it's correct...for 2...x^x^x=16 and x^(x^2)=16 try that of 3 also...and y=x^x^x means...y={[(x)^x]^x} and from indices u=(x^a)^b=x^ab...then y=x^(x.x)=x^(x^2)...so from the question y=x^x^x...taken natural log of both sides...lny=ln.x^x^x...lny=xlnx^x...(lnx^x=xlnx)then lny=x.xlnx...lny=x^2lnx...1/y.dy/dx=x+2xlnx...dy/dx=x^x^x{x+2xlnx}

Hi bro Calculusf(x), sorry, I have to disagree with you here.

x^x^x is not equal to (x^x)^x, but rather x^(x^x).

I am sorry if you haven't looked at it this way.

Please, look for your dy/dx again. Cheers!

smurfy:

Three years? Are you siwious?

As for being a genius, you say that because you're new to the thread. I wonder what you'll say when you see the real geniuses. I'm not kidding.

Note the correction to the original solution. It's +9 on the RHS, not -9.

smurfy: [/quote]

smurfy:

smurfy:

[quote author=smurfy]

smurfy:

Three years? Are you siwious?

Note the correction to the original solution. It's +9 on the RHS, not -9.

pls hw did u get +9 . I taught diminishes means _9

x + y + z = 3
x^2 + y ^2 + z^2 = 3
x^3 + y^3 + z^3 = 3

workings pliz!!

Humphrey77: maths guru please what is the solution to the equation 6"x-31(2"x)+32 equal to zero . (.please note the symbol " means raise to power)

.snap dis question and post.. Nt 2 clear.

Profmaojo: pls hw did u get +9 . I taught diminishes means _9

Let's look at the number itself, 65. 65 when interchanged (56) reduces it by 9. So, for 56 to 'catch up' with 65, it needs a +9, i.e. 65 = 56 + 9. You can see that this agrees with the argument in my solution.

You can also say 65 - 9 = 56, i.e. 65 needs to lose 9 'points' in order to 'catch down' with 56.

Catch down?

Calculusf(x):
.i respect you bro,but here is my approach...y=x^x^x...y=x^(x^2).try this with calculator and substitute values like 2 and 3 and check if it's correct...for 2...x^x^x=16 and x^(x^2)=16 try that of 3 also...and y=x^x^x means...y={[(x)^x]^x} and from indices u=(x^a)^b=x^ab...then y=x^(x.x)=x^(x^2)...so from the question y=x^x^x...taken natural log of both sides...lny=ln.x^x^x...lny=xlnx^x...(lnx^x=xlnx)then lny=x.xlnx...lny=x^2lnx...1/y.dy/dx=x+2xlnx...dy/dx=x^x^x{x+2xlnx}

y=x^x^x...y=x^(x^2).... wrong... I already posted a solution for this!

Humphrey77: OBIVIOUSLY 4-1 -is 3 .

we know that: (4-1)raise to power x + (4)raise to power x equals (4+1 ) raise to power x
clearly WE HAVE( 4(1-1/4))raise to power x + (4)raise to power x equal to (4(1+1/4) raise to power x . by linear approximation (1+x) raise to power n we have 1+


nx . when x is very small ( less than 1) clearly we have

4raise to power x( 1-x/4) + 4raise to power x equal to 4 raise to power x(1+x/4) so clearly we obtain the algebraic equation : 1-x/4 +1 is equal to 1+x/4 clearly we obtain
2-x/4 is equal to 1 + x/4
collect like term we have x equal to 2

thanks, buh I couldn't understand how u typed ur maths. buh I got sometyn "linear approximation"... will try it. thanks

Find the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

Humphrey77: find the missing numbers : 2,_10,_40 ( hint using commutative law of addition)

2, 8, 10, 20, 40..

smurfy: Find the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

5 x 6 x 7 x 8 + 3 = 1683

hardedeji1:


(5 x 6 x 7 x 8 ) + 3 = 1683

Correct! A little explanation would've been nice though.

Humphrey77: maths guru please what is the solution to the equation 6"x-31(2"x)+32 equal to zero . (.please note the symbol " means raise to power) ie 6 raise to power x -31(2 raise to power x) +32 equal to zero

use ur "linear approximation"
(1 + x)^n = 1 + nx

6^x - 31(2^x) + 32 = 0
(1 + 5)^x - 31((1 + 1)^x) + 32 = 0
1 + 5x - 31(1 + x) + 32 = 0
26x = 2
x = 1/13

if ur ansa doesn't give a close value to zero.... u can further try d binomial expansion...
i.e
(1 + x)^n = 1 + nx + [n(n - 1)x²]/x!

hardedeji1: x + y + z = 3
x^2 + y ^2 + z^2 = 3
x^3 + y^3 + z^3 = 3

workings pliz!!

answer =(1 ,1 ,1) ...from (1) square both sides.
We have (x+y+z)^2 =9
but, (x+y+z)^2 =x^2 + y^2 +z^2 + 2(xy+ yz+ zx)

==>9=3 + 2(xy+ yz + zx)
==> xy+ yz + zx=3 .....(4)
we should know that,
x^3 + y^3 + z^3 -3xyz=(x+y+z)(x^2 + y^2 + z^2 -xy-yz-zx)
==>3-3xyz=(3)(3-3)
which produce xyz=1.....(5)

now, let's have
x+y+z=3 . . . . .(1)
xy+yz+zx=3 . . .(4)
xyz=1 . . . .(5)

now, from (5)
yz=1/x and y+z=3-x (from (1)) pluck these in (4), we have 1/x + x(3-x) =3
...=> x^3 + -3x^2 +3x -1=0
by factor theorem , x=1trice
put these in (1) & (2)
we have
y+z=2 . . . .(6)
y^2 +z^2= 2 . . . .(7)
by substitution method
we have
y=1twice
==>z=1 (since z=2-1)

hence (1 , 1 , 1)

Calculusf(x):
.i respect you bro,but here is my approach...y=x^x^x...y=x^(x^2).try this with calculator and substitute values like 2 and 3 and check if it's correct...for 2...x^x^x=16 and x^(x^2)=16 try that of 3 also...and y=x^x^x means...y={[(x)^x]^x} and from indices u=(x^a)^b=x^ab...then y=x^(x.x)=x^(x^2)...so from the question y=x^x^x...taken natural log of both sides...lny=ln.x^x^x...lny=xlnx^x...(lnx^x=xlnx)then lny=x.xlnx...lny=x^2lnx...1/y.dy/dx=x+2xlnx...dy/dx=x^x^x{x+2xlnx}

hmmmm.
here is my skelewu approach.
y=x^x^x.
which implies y=x^(x^x)
recall dat y=x^x
log bth sides we av
lny=lnx^x.... lny=xlnx....
d/dy(lny)=lnx+1
1/y(dy/dx)=(lnx+1)
dy/dx=y(lnx+1) buh y=x^x
dy/dx=x^x(lnx+1).
now dat we knw dy/dx of y=x^x we can easily get y=x^(x^x)
log both sides we av
lny=lnx^(x^x)
lny=x^xlnx
d/dy(lny)=vdu/dx+udv/dx where my u=x nd v=x^x.
dy/dx(1/y)=x^x(1)+x(x^x(lnx+1))
dy/dx=y[x^x+x(x^x( lnx+1))]
recall dat y=x^x^x
finally we have
dy/dx=x^x^x[x^x+x(x^x(lnx+1))] shikena.

benbuks: answer =(1 ,1 ,1) ...from (1) square both sides.
We have (x+y+z)^2 =9
but, (x+y+z)^2 =x^2 + y^2 +z^2 + 2(xy+ yz+ zx)

==>9=3 + 2(xy+ yz + zx)
==> xy+ yz + zx=3 .....(4)

remainding solution loading

benbuks: answer =(1 ,1 ,1) ...from (1) square both sides.
We have (x+y+z)^2 =9
but, (x+y+z)^2 =x^2 + y^2 +z^2 + 2(xy+ yz+ zx)

==>9=3 + 2(xy+ yz + zx)
==> xy+ yz + zx=3 .....(4)

remainding solution loading

Lool, abeg drink water small b4 u resume o

hardedeji1:

2, 8, 10, 20, 40..

GOOD

Am taking breakfast now,.brb.

hardedeji1:


Lool, abeg drink water small b4 u resume o

hardedeji1:


Lool, abeg drink water small b4 u resume o

hardedeji1:


use ur "linear approximation"
(1 + x)^n = 1 + nx

6^x - 31(2^x) + 32 = 0
(1 + 5)^x - 31((1 + 1)^x) + 32 = 0
1 + 5x - 31(1 + x) + 32 = 0
26x = 2
x = 1/13

if ur ansa doesn't give a close value to zero.... u can further try d binomial expansion...
i.e
(1 + x)^n = 1 + nx + [n(n - 1)x²]/x!

NOTE PLEASE APPLYING LINEAR APPROX U NEED TO BE CARE FUL clearly, we know that (1+x)raise to power n is equal to (1+nx) for x is a small value like delta . example;: if we have 3 raise to power x using linear approx we have (4-1)raise to power x so factorize 4 we have( 4(1+1/4)) raise to power x clearly compare (1+x) and (1+1/4) you observe that x is 0.25 by comparison ( FOR ALL X IS SMALL)

rhydex 247:
hmmmm.
here is my skelewu approach.
y=x^x^x.
which implies y=x^(x^x)
recall dat y=x^x
log bth sides we av
lny=lnx^x.... lny=xlnx....
d/dy(lny)=lnx+1
1/y(dy/dx)=(lnx+1)
dy/dx=y(lnx+1) buh y=x^x
dy/dx=x^x(lnx+1).
now dat we knw dy/dx of y=x^x we can easily get y=x^(x^x)
log both sides we av
lny=lnx^(x^x)
lny=x^xlnx
d/dy(lny)=vdu/dx+udv/dx where my u=x nd v=x^x.
dy/dx(1/y)=x^x(1)+x(x^x(lnx+1))
dy/dx=y[x^x+x(x^x( lnx+1))]
recall dat y=x^x^x
finally we have
dy/dx=x^x^x[x^x+x(x^x(lnx+1))] shikena.

this particular line is whr u got it wrong...


dy/dx(1/y)=x^x(1)+x(x^x(lnx+1))

differential of x^xlnx should b

(x^x)/x + lnx[(x^x)(lnx + 1)]

recheck ur workings, note dat d/dx(lnx) is 1/x and not 1 as u wrote.

In how many different ways can the letters P, Q, R, S, T be arranged if the letter R must be to the right of the letter S?

**Goes out to buy gift for whoever thrashes the question**

Humphrey77: NOTE PLEASE APPLYING LINEAR APPROX U NEED TO BE CARE FUL clearly, we know that (1+x)raise to power n is equal to (1+nx) for x is a small value like delta

yea, that's y it is safer to use d binomial expansion to at least the 2nd degree
.
....