Please someone should help me solve these questions. My niece brought it to me but..............
1. x + y + z= 1
x² +y² + z²= 35
x³ + y³ + z³= 97

2.3^x + 4^x = 5^x

3. 3^x - 2^x = 65

4. 2log5₃ - 5log5₃
Please na beg i beg o

Question 1=later 2day
Question 2, recheck
Question 3= obviously X=65
Question4= -6.2907
Question 4 Soln
~2Log5^3 - 5Log5^3
2Log5^3=2X3Log5=6Log5
5Log5^3=5X3Log5=15Log5
therefore, the Qtn can b represented in simplified form as:
6Log5 - 15Log5= -9Log5.
Log5=0.6989(from tables)
-9Log5= -9 X 0.6989=
-6.2907....QED

Chimex388 thanks for help.
i know that in
2) x=2
3) x=4
but i dont just know how to table out my workings.
I get what you did in que 4 but the "3" is the base o.
I appreciate.

eph12: Chimex388 thanks for help.
i know that in
2) x=2
3) x=4
but i dont just know how to table out my workings.
I get what you did in que 4 but the "3" is the base o.
I appreciate.

u didnt type the qtn 2 properly(attachin the unknown 'x' to 4 ....
If my guess is correct, it shld've been:
3X + 4=5X
u didnt type Qtn 3 correctly as well if d ansa is 3...cus its equated to 65.....?
Thought the "3" in Qtn 4 was a power. back to work

Abeg no vex if it was as you think it is I would have done it since na. The x in 2 and 3 are powers o. Put the answers I supplied as the powers in place of x and you see am correct. I don't just know how to solve it so my niece understands. It was objective now it won't be a problem.

eph12: Please someone should help me solve these questions. My niece brought it to me but..............
1. x + y + z= 1
x² +y² + z²= 35
x³ + y³ + z³= 97

2.3^x + 4^x = 5^x

3. 3^x - 2^x = 65

4. 2log5₃ - 5log5₃
Please na beg i beg o

answer to question 1,
x=5, y=-1, z=-3

answer to question 2,
x=2

answer to question 3,
x=4

dejt4u:

answer to question 1,
x=5, y=-1, z=-3

answer to question 2,
x=2

answer to question 3,
x=4

Brother please help me with the calculations sir. You can just snap it and submit as attachment here. Thank you.

Freiburger: https://www.nairaland.com/1834634/need-mathematics-questions

Q1
. x + y + z= 1 .........................(B)

x² +y² + z² 35...........(E)

x ³+ y³ + z³= 97............(N)


SOLUTION :

From (E)

we observe that
(x+y+z)²= x²+ y² + z² + 2 (xy + xz + yz)

=> (1)²=35 +2(xy +xz+yz)

=> xy +xz + yz = -17 ..... ........(J)

also ,

(x+y+z) (x²+ y² +z² -xy-xz - yz )=x³ +y³ +z³ -3xyz

=> (1)(35 +17)=97-3xyz

=>3xyz=45
=>xyz=15............ ................(A)

Thus lets have
x+y+z=1.................................(B)
xy+xz+ yz =-17.....................(J)
xyz=15...................................(N)


From (B) , x+y= 1-z .................(M)
From (N), xy =15/z....................(I)

put these in (J)

=> 15/z + z(1-z) = -17

=> 15+z²-z³ = -17z

=> z³-z²-17z-15 =0

guess you can finish up from there ....( apply factor theorem or otherwise ) .

benbuks I appreciate. Please help with the workings for 2, 3 and 4. Thanks brother.

eph12: benbuks I appreciate. Please help with the workings for 2, 3 and 4. Thanks brother.

Q2

SOLUTION
lets use binomials

3^x =(1+2)^x

=>
3^x=1+2x+(4x(x-1))/2 +…

=> 3^x =1+2x² +…

similarly ,
4^x=(2-3x+9x² ) /2 + …
&
5^x =1-4x+8x² + …

recall that
3^x +4^x=5^x

=> 1-4x+8x² +…=1+2x² +...+(2-3x+9x² )/2 +…

regrouping identical terms & factorizing

=> 3x²-5x-2=0

thus x=2 or -1/3

hence x=2

benbuks:

Q2

SOLUTION
lets use binomials

3^x =(1+2)^x

=>
3^x=1+2x+(4x(x-1))/2 +…

=> 3^x =1+2x² +…

similarly ,
4^x=(2-3x+9x² ) /2 + …
&
5^x =1-4x+8x² + …

recall that
3^x +4^x=5^x

=> 1-4x+8x² +…=1+2x² +...+(2-3x+9x² )/2 +…

regrouping identical terms & factorizing

=> 3x²-5x-2=0

thus x=2 or -1/3

hence x=2

Great! You have been of much help. Just tell me what 2^x is I can continue from there. Thanks bro.

eph12:
Great! You have been of much help. Just tell me what 2^x is I can continue from there. Thanks bro.

come in here www.nairaland.com/1147658/nairaland-mathematics-clinic/140#25103637

lets continue .

eph12:
Chimex388 thanks for help.
i know that in
2) x=2
3) x=4
but i dont just know how to table out my workings.
I get what you did in que 4 but the "3" is the base o.
I appreciate.

i can solve 2 and 3 and snap via whatsapp for you