Question

Determinati numarul natural x care verifica egalitatile:
a)5^x+2×2^x+1+7×10^x=57000
b)12×4^x-5×4^x=7×2¹²
c)4^x+2+4^x+1+4^x=336
d)9⁴³-9⁴²-9⁴¹=71×3^x
e)7^x+3-7^x+3-7^x+1=41×7^2×x
f)3^x+3×4^x+1-5×3^×4^ x=1236

Answer 1

a)5^x+2×2^x+1+7×10^x=57000
5^x+2×2^x+1+7×5^x×2^x=57000
5^x×2^x(5^2×2^1+7)=57000
5^x×2^x ×57=57000
5^x×2^x =1000
10^x =10³
x=3

b)12×4^x-5×4^x=7×2¹²
12×2^2x-5×2^2x=7×2¹²
2^2x(12-5)=7×2¹²
2^2x ×7=7×2¹²
2^2x =2¹²
x=12

c)4^x+2+4^x+1+4^x=336
2^2x+4+2^2x+2+2^2=2^4 × 21
2^2x(2^4+2^2+2^0)=2^4 × 21
2^2x(16+4+1)=2^4 × 21
2^2x ×21 =2^4 × 21
2^2x =2^4
2x=4
x=2

d) 9⁴³-9⁴²-9⁴¹=71×3^x
3⁸⁶-3⁸⁴-3⁸²=71×3^x
3⁸²(3⁴-3²-3⁰)=71×3^x
3⁸²(81-9-1)=71×3^x
3⁸² ×71=71×3^x
3⁸²=3^x
x=82

e) aici cred ca termenul al 2-lea este 7^x+2
7^x+3-7^x+2-7^x+1=41×7^2×x
7^x+1(7^2-7^1+3-7^0=41×7^2×x
7^x+1(49-7-1)=41×7^2×x
7^x+1× 41=41×7^2×x
x+1=2x
x=1