How do you solve 1/16=64^(5x-4)?

thank you

Notice that 16 and 64 are both powers of 2.

2^-4 = 2^6(5x-4)

take log base 2 of each side..

-4=6(5x-4)
solve for x

wow, okay thanks!!!

so its -.93?

To solve the equation 1/16 = 64^(5x-4), follow these steps:

1. Notice that 16 can be expressed as 2^4 and 64 can be expressed as 2^6. Rewrite the equation using these powers of 2:
2^(-4) = (2^6)^(5x-4)

2. Apply the power rule in exponents: (a^m)^n = a^(m*n):
2^(-4) = 2^(6*(5x-4))

3. Now, we have 2^(-4) on one side and 2^(6*(5x-4)) on the other side. We can equate the exponents of 2 to solve for x.

4. Take the logarithm base 2 of both sides. This will help us remove the exponent and solve for x:
log2(2^(-4)) = log2(2^(6*(5x-4)))

5. According to the logarithm property log2(a^m) = m*log2(a), we can rewrite the equation as:
-4*log2(2) = (6*(5x-4))*log2(2)

6. Simplify the equation using the fact that log2(2) equals 1:
-4 = 6*(5x-4)

7. Distribute 6 to both terms in parentheses:
-4 = 30x - 24

8. Rearrange the equation:
30x - 24 = -4

9. Add 24 to both sides:
30x = 20

10. Divide both sides by 30:
x = 20/30

11. Simplify:
x = 2/3

Therefore, the solution to the equation 1/16 = 64^(5x-4) is x = 2/3.