Question

if 2352=2^x*3^y*7^z then the value of x+y+z

Answer 1

Begin by splitting the number into its prime factors,

2352 = 2 * 2 * 2 * 2 * 3 * 7 * 7
=> 2352 = 2 ^ 4 * 3 ^ 1 * 7 ^ 2

Also according to the question,

2352 = 2 ^ x * 3 ^ y * 7 ^ z

Hence we can conclude that,

2 ^ 4 * 3 ^ 1 * 7 ^ 2 = 2 ^ x * 3 ^ y * 7 ^ z

Hence,

x = 4, y = 1 and z = 2

Now adding the values of x, y and z,

x + y + z = 4 + 1 + 2
=> x + y + z = 7

Answer 2

consider LHS:
2352 =2^4*3^1*7^2
now consider RHS:
2^x*3^y*7^z =LHS
Through LHS and RHS,
2^4*3^1*7^2 = 2^x*3^y*7^z
== x=4
y=1
z=2
therefore , x+y+z = 4+1+2
=7
I wish my answer is helpful for you