Question

If x^3 < 8^4, then find the greatest possible integral value of x with step by step explanation

Answer 1

Step-by-step explanation:

8⁴=8*8*8*8

8⁴=8*8*8*8 =512*8

8⁴=8*8*8*8 =512*8 =4096

8⁴=8*8*8*8 =512*8 =4096Now....

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 1

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 2

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8.

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8..

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8...

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8....

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8....possibility (15),

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8....possibility (15),when x = 15,

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8....possibility (15),when x = 15,15³= 3375

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8....possibility (15),when x = 15,15³= 3375possibility (16),

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8....possibility (15),when x = 15,15³= 3375possibility (16),when x = 16,

8⁴=8*8*8*8 =512*8 =4096Now....possibilities of x³…possibility (1),when x= 11³=1possibility (2),when x= 22³=8....possibility (15),when x = 15,15³= 3375possibility (16),when x = 16,16³=4096

now..,

now..,possibility 16 is equal to the value of 8⁴...

now..,possibility 16 is equal to the value of 8⁴...so,

now..,possibility 16 is equal to the value of 8⁴...so,the greatest integral value of x³ is possibility (15).

now..,possibility 16 is equal to the value of 8⁴...so,the greatest integral value of x³ is possibility (15).that is ....x= 15 is the correct answer...

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Answer 2

Answer:

If x3<84, then find the greatest possible integer value of x.

The greatest possible satisfying the above inequation is 15.