How do you solve log_64 y<=1/2?

Nov 3, 2016

Explanation:

Before we begin, add the restriction, $y > 0$ because the argument for a logarithm can never be less than or equal to zero:

log_64(y) <= 1/2; y > 0

To make the logarithm disappear, write both sides as exponents of 64:

64^(log_64(y)) <= 64^(1/2); y > 0

The left side reduces to y and we can drop the restriction if we add $0 < y$ to the expression:

$0 < y \le {64}^{\frac{1}{2}}$

The square root is the same as the $\frac{1}{2}$ power:

$0 < y \le \sqrt{64}$

$0 < y \le 8$