₦airaland Forum

Sir-Tunechi:
Mathz generalz take diz one chil
1^x + 2^x =3^x-1
find x
p.s:i nor wan hear newton raphson method o

let # denot congruency....then
1+2^x#0 (mod3) or
2^x#-1 (mod3) or 2^x#2 (mod3)
or 2^(x-1)#1 (mod3) by Fermat's Little Theorem, x-1 must divide @(3)=3-1=2, Euler's phi function. So x-1=1 or 2 so that x=2 or 3...but x=2 does not satisfy d given eqn so x=3....

Mbahchiboy: i wud like to ask d generals in dis thread
''y are u all been so adamant towards my questions

almost everybody gave solutions to your questions on the previous pages...unless u have not viewed them....

Mbahchiboy: MAKE UNA HELP ME OUT O........
**On my knees**

for your first question use Maclaurin's series...the solution is;
tanhx=x-2x^3/3!+4x^5/5!-...+...
(-2)^n*x^(2n+1)/(2n+1)! For
n=0,1,2...
For your second question,
x^5-32=0 may be written as
x^5=32e^(2#i) where #=22/7 and e=2.71828, it follows that
x=2e^(2#ni/5) for n=0,1,2,3,4 they provide all solutions to the equation...for other soutions check preivous pages...

rhydex 247: Breakfast questn
1. Solve. (x^2-16)(x-3)^2+9x^2=0.

@Rhydex...here's a simple approach...rearrang d eqn to take d followin form...
9x^2=(16-x^2)(x-3)^2...inspection shows that the first expression on the right must be a perfect square...16-x^2=y^2.......eqn1 so we get 9x^2=y^2*(x-3)^2 or
3x=xy-3y or xy=3(x+y)........eqn2 now from eqn1 x^2+y^2=16 or (x+y)^2-2xy=16 hence we can substitut for xy from eqn2 to get;
(x+y)^2-6(x+y)-16=0 so that
x+y=8 or -2 then xy=24 or -6 we now sovlv quadratically for x & y;
z^2-(x+y)z+xy=0 or z^2-8z+24=0 & z^2+2z-6=0...solvin the above quadratics gives u d four values of x...thus x=......plz i have no calculator on me now...

rhydex 247: @ Sir tunechi questn.
1^x+2^x=3^(x-1).
solution.
take log of both sides.
log1^x+log2^x=log3^(x-1).
xlog1+xlog2=(x-1)log3.
xlog1+xlog2=xlog3-log3.
recall that log1=0, log2=0.3010 nd log3=0.4771.
Nw we have.
0.3010x=0.4771x-0.4771.
0.4771=0.4771x-0.3010x.
0.4771=0.17612x
x=0.17612/0.4771
x=2.708929145.
Approximately
x=3.
I want to ask wetin newton raphsön method do u @ sir tunechi.

@Rhydex...i dont think d Log of a sum is d same as d sum of d log...as an example...Log(2+3) is not equal to Log2+Log3...if u stil insist, here is a mathematical proof...let a & b any two positive numbers then
Log(a+b)=Loga +Logb now from logarithm, Logab=Loga+Logb...so that Log(a+b)=Logab or a+b=ab that is, d sum of any two numbers is always equal to their product which is blatantly false...

Sir-Tunechi:
Mathz generalz take diz one chil
1^x + 2^x =3^x-1
find x
p.s:i nor wan hear newton raphson method o

we divide thru bt 3^x ; 1/3^x+(2/3)^x=1/3, now let a=1/3^x then (2/3)^x=1/3-a now from d above parametric equations, xLog(1/3)=Loga & xLog(2/3)=Log(1/3-a)...dividin d last eqn by d precedin one gives, 0.369=Log(1/3-a)/Loga or a^0.369=1/3-a...takin cube of both side; a=(1/3-a)^3 approx. let b=1/3-a, then last eqn becoms 1/3-b=b^3 or b^3+b-1/3=0...solve for b, then for a and then for x from a=1/3-b, 1/3^x=a, u 'll find that x=3 approx...sorry i have no calculator on me...

rhydex 247: @ laplacian question. Solutiön.
Abstract Algebra
The center is a subgroup. We show closure nd inverses. Suppose x,y in Z(G) implies xa=ax and ya=ay for every a in G. Nw xa=ax implies x=axa^-1 so that. xya=(axa^-1)ya=axy. Since nw that (xy)a=a(xy), by definition xy lies in Z(G). [we av used inverses because all of the elements are in G and hence av inverses].
nw we show that if x is in Z(G), then x^-1 lies in Z(G). If x is in Z(G) then xa=ax for all a in G. Then xa=ax implies a=x^-1ax => ax^-1=x^-1a. Which means x^-1 is in Z(G). Nw dat we av showed Z(G) is a subgroup of G. We nw show dat it is normal in G. Let H=Z(G). We must show that H=yHy^-1 for all y in G. Let x be in H. Then we know that yx=xy but yxy^-1=xyy^-1=x. Hence x is in yHy^-1. Suppse x is in yHy^-1. Then x=xyy^-1=yxy^-1 so dat xy=yx and hence x is in H. Hence the center of a group Z(G) is a normal subgroup of G.

thumbs up to u, @Ryhdex

...show that the center of a group Z(G) is a normal subgroup of G....