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#