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

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Answer 1 · 2016-11-03T19:43:20

Please see the 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:

$0 < y <= 64^(1/2)$

The square root is the same as the $1/2$ power:

$0 < y <= sqrt64$

$0 < y <= 8$

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