emmyeuler1: general double dx i hail ooo...its bin a long time

emmyeuler1: i have two solutions but i'll start wit d easier one. Let # denote "congruency" then since 3^x+4^x=5^x, 4^x#5^x(mod3) or by Fermat's Little theorem (4^x)^2#(5^x)^2#1(mod3), i.e 4^2x#5^2x#1(mod3) or 16^x#1(mod3) and 25^x#1(mod3) so by FLT, x must divide @(3)=3-1=2 Euler's phi function, i.e x|2 hence either x=1 or 2 but since 1 does not satisfy the eqn, x=2,

Odehfamily: 5/12 of a number is subtracted from 3/4 of the number. The positive difference is 7 less than 5/6 of the number. Find the number. plz help me

Odehfamily: 5/12 of a number is subtracted from 3/4 of the number. The positive difference is 7 less than 5/6 of the number. Find the number. plz help me

busuyem:

Still expecting the solution pls.

busuyem: Pls, help me with the solution of this question with its explanation:

The difference between the digits of a two-digit number is 1. The number itself is 1 more than 5 times the sum of its digits. If the units digit is greater than the tens digit, find the number.

Calculusf(x):
...y=∫√(1+√x)/x.dx...let √x=a²...x=a^4...dx=4a^3.da...then,∫√(1+a²)/a^4.4a^3da...∫√(1+a²)/a..da...if u²=1+a²...udu=ada...da=udu/a...then ∫u/a*udu/a...∫u²/a².du but u²=1+a²...a²=u²-1...then ∫u²/(u²-1).du...do the division to get...∫1+1/(u²-1).du...from here...let's solve 1/u²-1...resolve to partial fraction,1/u²-1=(A/u-1)+(B/u+1)....1=A(u+1)+B(u-1)...if u=1...A=1/2...and B=-1/2....which gives,1/2(u-1)-1/2(u+1)...then,∫1+1/2(u-1)-1/2(u+1).du...then,{u+1/2ln(u-1)-1/2ln(u+1}...{u+1/2ln(u-1)/(u+1)}+c...don't forget that u²=1+a²...u=√1+a²...don't forget that a²=√x...u=√1+√x...substitute that to the answer to give...{√1+√x+(1/2)ln(√1+√x-1)/(√1+√x+1)+c...{√1+√x+(1/2)ln(x^1/4)/(√2+√x)+c...i guess.....waiting for better solutions from my generals

Eng. Tino:
choii...........all dese ones are uni maths....i hail oo.....i av a question.......a four digit positive number 'N' is given.....when d tens digit and unit digit of d numba is swapped and the thousand digit and hundred digit is also swapped, a new number T(N) is formed.... M is a number such dat T(M)=4M................what are d last 3 digits of M

Ortarico:

Even nondeterministic polynomial (NP) method can be used.
x = x1 - f(x1)/f'(x1)

3^x + 4^x = 5^x
3^x + 4^x - 5^x = 0
f(x) = 3^x + 4^x - 5^x
from polynomial:
f(x1) = 3^x + 4^x - 5^x
let x1 = 2
f(2) = 3^2 + 4^2 - 5^2
f(2) = 9 + 16 - 25
f(2) = 0
:. x = 2 is a factor

f'(x) = 3^xIn3 + 4^xIn4 - 5^xIn5
f'(2) = 9In3 + 16In4 - 25In5
f'(2) = 9.89 + 22.18 - 40.24
f'(2) = -8.17

i.e x1 = 2, f(2) = 0 and f'(2) = -8.17. . . .substituting into the formula above:
x = 2 - 0/-8.17
x = 2 + 0
x = 2

Calculusf(x):
...great master but i guess this is newton-raphson iterative method

busuyem:

No solution yet?

Calculusf(x):
...let the number be xy...from the question,we were told that the difference is 1,and the unit digit is greater than the ten digit,which means y-x=1...then y=1+x...EQUATION 1....the question states again that the original question is one more than 5 times the sum of the digits...from xy...the digits are x and y but the original one is 10x+y...assuming 23 which will equal to 10*2+3...then 10x+y=5(x+y)+1...10x+y=5x+5y+1....5x-4y=1...don't forget that y=1+x from equation 1,substitute that to give 5x-4(1+x)=1...5x-4x-4=1...therefore x=5...from equation 1...y=x+1=5+1 to give 6...since the number is xy...then the number is 56...

d citizen:
simplify sqr (6 +sqr (6 + sqr(6+ to infinite)))

Note sqr mean square root.

d citizen:
Solve 4^x = 8x

d citizen:
@emmyeuler1 ur solution to the problem 4^x = 8x show dat u re no a math general but a math major. Look log2(2x -3) = (2x-3)log 2

Or log2 (2x -3) =log2^(2x-3) stop dat ur wuru to the answer. Your solution to the problem is wrong.