How do you solve ${log}_{64} y \leq \frac{1}{2}$ $log_64 y<=1/2$ ?

Question

1758 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-03T19:43:20

Please see the explanation

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

${log}_{64} (y) \leq \frac{1}{2}; y > 0$ $log_64(y) <= 1/2; y > 0$

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

$64^{{log}_{64} (y)} \leq 64^{\frac{1}{2}}; y > 0$ $64^(log_64(y)) <= 64^(1/2); y > 0$

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

$0 < y \leq 64^{\frac{1}{2}}$ $0 < y <= 64^(1/2)$

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

$0 < y \leq \sqrt{64}$ $0 < y <= sqrt64$

$0 < y \leq 8$ $0 < y <= 8$

How do you solve log64y≤12log_64 y<=1/2?