Tìm x$\in$N, biết.

a, $2^x.2^2=32$

b, $27.3^x=243$

c, $2^x.2^4=1024$

d, $49.7^x=2401$

a, $\left(2^{17}+{17}^2\right)\left(9^{15}-3^{15}\right)\left(2^4-4^2\right)$

= $\left(2^{17}+{17}^2\right)\left(9^{15}-3^{15}\right)\left(16-16\right)$

= $\left(2^{17}+{17}^2\right)\left(9^{15}-3^{15}\right).0$

= 0

b, $\left(1^2+2^3+3^4+4^5\right)$.$(1^3+2^3+3^3+4^3)(3^8-{81}^2)$

= $\left(1^2+2^3+3^4+4^5\right)$.$(1^3+2^3+3^3+4^3)(3^{4.2}-{81}^2)$

= $\left(1^2+2^3+3^4+4^5\right)$.$(1^3+2^3+3^3+4^3)({81}^2-{81}^2)$

= $\left(1^2+2^3+3^4+4^5\right)$.$(1^3+2^3+3^3+4^3).0$

= 0

c, $\left(7^{24}+7^{23}\right):7^{22}$

= $7^{24}:7^{22}+7^{23}:7^{22}$

= $7^{24-22}+7^{23-22}$

= $7^2+7^1$

= 49 + 7 = 56

a, $3^{x+1}:3^4=81$

$3^{x-3}=3^4$

x – 3 = 4

x = 7

Vậy x = 7

b, $3^{x+3}.3^{x+1}=729$

$3^{2x+4}=3^6$

2x + 4 = 6

x = 1

Vậy x = 1

c, $2^{x+3}.2^x=128$

$2^{2x+3}=2^7$

2x + 3 = 7

x = 2

Vậy x = 2

d, $23+3x=5^6:5^3$

$23+3x=5^3$

23 + 3x = 125

3x = 102

x = 34

Vậy x = 34

e, $2^x+2^{x+4}=272$

$2^x+2^x.2^4=272$

$2^x(1+2^4)=272$

$2^x.17=272$

$2^x=16$

$2^x=2^4$

x = 4

Vậy x = 4