Question

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

Answer 1

`7^(log_7 x) = x`

with restrictions on the value of `x`

Explanation:

By the very definition of logarithm, for any valid base `b` (i.e. `b > 0` and `b != 1`):

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

So if `log_b x` exists then by definition, `b^(log_b x) = x`

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Real valued logarithm

If we restrict ourselves to Real values of `y`, then note that:

`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 `y |-> b^y` is continuous and strictly monotonic increasing, being a one-one function from `(-oo, oo)` onto `(0, oo)`.

So if `x > 0` then there is a unique solution `y` of `b^y = x` and so `log_b x` is defined.

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Complex valued logarithm

If we allow Complex values of `y`, then note that:

`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 `y -> b^y` is a continuous many to one function from `CC` onto `CC "\" { 0 }`

If `b^y = x` then `b^(y + (2npii)/(ln b)) = x` for any integer value of `n`

So for `x != 0` we can pick a principal value of `log_b x` such that `Im (log_b x) in (-pi/(ln b), pi/(ln b)]`

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 `log_b x`, by definition we still have:

`b^(log_b x) = x`