+1 vote
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A computer uses ternary system instead of the traditional systen, An $n$ bit string in the binary system will occupy

1. $3+n$ ternary digits
2. $2n/3$ ternary digits
3. $n$$\log_{2}3 ternary digits 4. n$$\log_{3}2$ ternary digits

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Let the no. of digits occupied by the ternary system be 'x'

Now,Maxi. value representable in ternary system using x digits = Maxi. value representable in binary system using  n digits

3x-1=2n-1

x=log32n = n log32   (d)

by Boss (10.8k points)
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Maximum No. possible using n-bit in Binary:- 2n -1

Maximum no. possible using x bit in Ternary:- 3x -1

Both will take different no. of bits to represent same number.

3x -1 =  2n -1

3=  2n

taking log on both side:;-

X= log3 ( 2 )

X=n*log32 .

by Active (3.3k points)
0
@Sonveer tomar 1,Could you elaborate how you've got 3 in base of log?What property you've used?
+1

For example:-

10n = 1000. For what value of n we get this

take log base 10 on both side.

log(10n ) = log(1000)

n log(10) = 3* log(10)

n=3.

In same manner to find value of X i take log base 3 on both side

0
@Sonveer tomar 1,Thank you so much for the explanation. :)
0
+1

the problem is simply trying to ask---> given an n-bit binary number howmany digits that that it would be required to represent that number in ternary number system....so here we try to find howmany number of digits in ternary number system that will be required to represent the maximum valued n-bit binary number......

so,let x digits required in ternary number system to represent the maximum valued n-bit binary number which is (2^n-1)(in decimal)....and maximum valued decimal number with x digit in ternary number system will be (3^x-1)...

so ,(2^n-1)<=(3^x-1)

=>  2^n<=3^x

=>  x>=nlog2base3