Question

How do you evaluate `log_625 25`?

Answer 1

`log_625 25 = color(green)(1/2)`

Explanation:

Note that `625 = 25^2color(white)("XX")rarr 25 = 625^(1/2)`

Therefore
`color(white)("XXX")log_625 25 = log_625 626^(1/2)`

`color(white)("XXX")rarr log_625 25 = 1/2` (by definition of log)

Answer 2

`log_625 25 = 1/2`

Explanation:

Logs are easier to understand if you think about an expression given in log form as asking a question.

In `log_625 25`, the question being asked is;

"What power/index of 625 will give 25?"
OR " How do I make 625 into 25?"

You should recognise `25` as being the square root of 625.

A square root can be written as an index.

`sqrtx = x^(1/2)`

`sqrt625 = 625^(1/2) larr` this answers our question!

`log_625 25 = 1/2`

Log form and index form are interchangeable.

`log_a b = c hArr a^c = b`

`log_625 25 = x hArr 625^x = 25`

`625^x = (25^2)^x= 25^1 larr` make the bases the same

`25^(2x) = 25^1 larr` equate the indices and solve for x

`2x = 1 rarr x = 1/2`