1/64=16^2a

4^-3=(2^4)^2a
4^-3=2^8a
-3=8a
-3/8=a

I don't understand what I'm doing incorrectly can some please explain to me,in detail

you cannot equate exponents unless the bases are the same

4^-3 = (4^2)^2a
4^-3 = 4^4a
-3 = 4a
a = -3/4

or

4^-3 = 16^2a
2^-6 = 2^8a
-6 = 8a
a = -3/4

I got the answer thank you

Sure, I can explain the steps to you.

Let's start with the first equation:

1/64 = 16^(2a).

To solve for 'a', we need to get rid of the exponent on the right side of the equation. To do that, we can take the logarithm (base 16) of both sides of the equation.

log(base 16)[1/64] = log(base 16)[16^(2a)]

Now, on the left side of the equation, the logarithm (base 16) and exponent (1/64) cancel out, as the logarithm and exponent are inverse operations.

0 = 2a

Next, divide both sides of the equation by 2 to isolate 'a'.

0/2 = 2a/2

0 = a

So, the solution to the first equation is 'a = 0'.

Let's move on to the second equation:

4^(-3) = (2^4)^(2a).

Again, we want to get rid of the exponent on the right side of the equation. To do that, we can simplify the exponent first.

On the right side, we have (2^4)^(2a), and according to the power of a power rule, we can multiply the exponents together:

(2^4)^(2a) = 2^(4 * 2a) = 2^(8a).

Now, the equation becomes:

4^(-3) = 2^(8a).

To compare the bases, we can take the logarithm (base 2) of both sides of the equation.

log(base 2)[4^(-3)] = log(base 2)[2^(8a)]

Using the power rule of logarithms, we can bring the exponent down to the front of the logarithm on the right side.

-3 * log(base 2)[4] = 8a * log(base 2)[2]

On the left side, we know that log(base 2)[4] is equal to 2 because 2^2 = 4.

-3 * 2 = 8a * log(base 2)[2]

-6 = 8a * 1

Now, divide both sides of the equation by 8 to solve for 'a'.

-6/8 = 8a/8

-3/4 = a

So, the solution to the second equation is 'a = -3/4'.

I hope this explanation helps you understand the process better. Let me know if you have any more questions!