Question

Sa se afle x din:
3^1*3^2*...*3^x-(3^2)^17=8*9^17

Answer 1

3 ^ (1+2+...+x)- 3^(2*17)=2^3 *3^ (2*17), x≥0
3^ x(x+1)/2 = 3^34(1+8)
3^x(x+1)/2=3^36
x(x+1)/2=36
x²+ x- 72=0
(x-8) (x+9)= 0
x₁= -9 <0 nu e bun
x₂= 8

O zi buna!

Answer 2

$3^1 \cdot 3^2\cdot...\cdot3^x-(3^2)^{17}=8 \cdot 9^{17} \\ 3^1 \cdot 3^2\cdot...\cdot3^x-9^{17}=8 \cdot 9^{17} \\ 3^1 \cdot 3^2\cdot...\cdot3^x = 8 \cdot 9^{17} + 9^{17} \\ 3^1 \cdot 3^2\cdot...\cdot3^x = 9 \cdot 9^{17} \\ 3^1 \cdot 3^2\cdot...\cdot3^x = 9^{18} \\ 3^1 \cdot 3^2\cdot...\cdot3^x = (3^2)^{18} \\ 3^1 \cdot 3^2\cdot...\cdot3^x = 3^{36} \\ 1 +2+...+x = 36 \\ x = 8$