# How do you simplify #7^(log_7x)#?

##### 1 Answer

with restrictions on the value of

#### Explanation:

By the very definition of logarithm, for any valid base

#y = log_b x" "# is a number such that#" "b^y = x#

So if

**Real valued logarithm**

If we restrict ourselves to Real values of

#b^y > 0" "# for all Real values of#y#

So for Real valued logarithms:

#log_b x# is only definable when#x > 0#

In fact, the function

So if

**Complex valued logarithm**

If we allow Complex values of

#b^y != 0" "# for all Complex values of#y#

So for Complex valued logarithms:

#log_b x# is only definable when#x != 0#

In fact, the function

If

So for

With that convention, we would have:

#log_b x = (ln x)/(ln b) = 1/(ln b) (ln abs(x) + Arg(x) i) = log_b abs(x) + (Arg(x))/(ln b) i#

Whichever convention we use to pick the principal value of

#b^(log_b x) = x#