Indices

2^x = 2^3(y+1)

x = 3y+3 —– ❶

9^y = 3^x-9

3^2y = 3^x-9

2y = x-9 —– ❷

x = 21, y = 6

Therefore, x^y = 21⁶ = (21²⁾³ = 441³

Let a^p=b^q=c^r=k

a = k^1/p, b = k^1/q, c = k^1/r

Now, c = ab (given)

Therefore, k^1/r = k^1/p. k^1/q

-> k^1/r = k^1/p+1/q

-> 1/r = 1/p + 1/q
-> r = pq/(p+q)

Given, 2^x = 4^y = 8^z
-> if x =1, then y =1/2 and z = 1/3
Now, the LCM of 2 and 3 (denominators of the above values of y and z) is 6 -> the first set of values of x, y and z if they are all natural numbers, will be 6, 3 and 2 respectively.
Thus, 2⁶ = 4³ = 8²
-> x+y+z = 11

Let x⁵ = y
Then x = y^1/5
Therefore, (y^1/5)^y = 5
-> y^y/5 = 5
-> y^y = 5⁵

-> y = 5
-> x⁵ = 5

-> x = 5^1/5

 

Alternatively

Check the value of x from the options, which satisfies the equation

Given, x^pqr = x^p.x^q.x^r = x^p+q+r

-> pqr = p+q+r

By trial and error, p, q, r are 1, 2, 3 (respective values are not important as each variable undergoes the same treatment).

-> 1³+2³+3³ = 36

Nine raised to the power 8^{7 =} 9^{8*8*8…..7 times}
9⁸ raised to the power 7 = (9⁸)⁷ = 9⁵⁶
A. Therefore, nine raised to the power 8⁷ is more than 9⁸ raised to the power 7.

A five-digit value of 3^x+4^y+5^z is only possible if z is 6 or 7.  This expression will not be a five-digit number for z values less than 6 and more than 7. However, since 3^x+4^y+5^z is the largest five-digit number, we take the value of z as 7.

Again, for 3^x+4^y+5^z to be maximum under the given conditions, x= 6 and y=7. For x=6, y=7, z=7, 3^x+4^y+5^z is still a five-digit number. Thus, plug in these values to find x⁴+y³+z².

Desired value = 6⁴ + 7³ + 7² = 1296 +343 +49 = 1688

p = q² —– ❶

q = p^q —– ❷

Substituting ❶ in ❷

q = (q²)^q = q^2q  -> q = 1 or ½

when q = 1, p = 1 -> Option C is possible

when q = ½, p = ¼

We check other options using the above values.

Option A is possible when p = ¼, and q = ½ as (1/4)² + (1/2)² = 5/16

Option B is possible when p = ¼ and q = ½ as (1/4)/(1/2)³ = 8/4 =2

Therefore, all of them might be true.

Given, yⁿ = z⁴
-> x raised to 2n⁴⁺¹ = x raised to 4ⁿ⁺¹
-> 2n⁴⁺¹ = 4ⁿ⁺¹
-> 2n⁵ = 2²ⁿ⁺²
-> 2n⁵ = 2.2²ⁿ⁺¹
-> n⁵ = 2²ⁿ⁺¹
By trial and error, we find that n = 2
-> ⁽²ⁿ⁺¹⁾√n²⁰ = ⁵√2²⁰ = (2²⁰)^1/5 = 2⁴ =16