Can someone help me with these algebra problems?

Question

1. (a^k)^k=a^9
2. (2^k)*(2^4)=2^7

Answer 1

(a^k)^k = a^9
a^(k^2) = a^9
k^2 = 9
k = 3 or -3

(2^k)*(2^4) = 2^7
2^(k + 4) = 2^7
k + 4 = 7
k = 3

Answer 2

1. a^(k^2)=a^9
k^2=9
k= +or- 3
2. 2^(k+4)=2^7
k+4=7
k=3

Answer 3

(a^k)^k is the same thing as a^(k*k) or a^(k^2).

If a^(k^2) = a^9 then k^2 = 9. Square root both sides and you get k = +/- 3. a can be any real number.

(2^k)*(2^4) is the same as 2^(k+4).

If 2^(k+4) = 2^7 then k+4 = 7. k = 3.

Answer 4

I kinda solved these logically, but i'll try to explain
If a^k to ^k, then you mulitply the k's, so k^2 = 9
thus k= 3, and -3

And then 2^K * 2^4 = 2^7
so divide 2^7 by 2^4 = 2^3
so 2^k = 2^3
so k = 3

Answer 5

let's start with equation (1)

it's of the form (x ^ a) ^ a = x ^(a times a)

so that in your case,

(a^k)^k = a ^ (k*k) = a ^ 9

or

k^2 = 9

and k = +/- 3


equation (2)

it is of the form (x ^ a) * (x ^ b) = x ^ c
remember, (x ^ a) * ( x^ b) = x ^ (a + b)

therefore a + b = c,
or in your case
k + 4 = 7 and k = 3

Answer 6

1. k=3 b/c (k)^2=(3)^2
2. k=3 b/c k+4=7 therefore k=3

I hope this helps!

Answer 7

When exponents are raised to another exponent, they are multiplied, so your first one is a^(k*k) = a^9 and k=3

When numbers with exponents are multiplied, the exponents are added. So the second is 2^(k + 4) = 2^7 and k=3

Answer 8

1. (a^k)^k = a^(k^2) = a^9

k^2 = 9

k = 3, -3


2. (2^k)*(2^4) = 2^7
2^(k+4) = 2^7

k+4 = 7
k = 3

Answer 9

1. (a^k)^k=a^9

a^(k²) = a^9
k² = 9
k = ±3

2. (2^k)*(2^4) = 2^7

2^(k+4) = 2^7
k = 4 = 7
k = 3

Answer 10

2. K=3
find the answer for 1 one