Question

How do you evaluate `log_64 8`?

Answer 1

`log_64 8=1/2`.

Explanation:

We will use the Defn. of `log` fun. to find the desired value.

Recall that, `log_ba=x iff b^x=a`.

So, `log_64 8=x rArr 64^x=8 rArr (8^2)^x=8 rArr 8^(2x)=8^1`.

As `log` fun. is `1-1`, we must have, `2x=1," giving, "x=1/2`.

`:. log_64 8=1/2`.

Answer 2

`log_(64)8=1/2`

Explanation:

When I write `log_ab=c`, I ask to what power I raise the base `a` to get `b`; the answer here is `c`, i.e. `a^c=b`. For `log_(10)10`, `log_(10)100`, `log_(10)1000`, `log_(10)1000000`, I can immediately write `1,2,3,6` without thinking too much.

For your problem, `log_(64)8`, I look for the power to which I raise `64` to get `8`. But of course, `8^2=64`, or `sqrt64=8`, `64^(1/2)=8`. So `log_(64)8=1/2`.