# How do you simplify 7^(log_7x)?

Oct 1, 2016

${7}^{{\log}_{7} x} = x$

with restrictions on the value of $x$

#### Explanation:

By the very definition of logarithm, for any valid base $b$ (i.e. $b > 0$ and $b \ne 1$):

$y = {\log}_{b} x \text{ }$ is a number such that $\text{ } {b}^{y} = x$

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:

${b}^{y} > 0 \text{ }$ for all Real values of $y$

So for Real valued logarithms:

${\log}_{b} x$ is only definable when $x > 0$

In fact, the function $y \mapsto {b}^{y}$ is continuous and strictly monotonic increasing, being a one-one function from $\left(- \infty , \infty\right)$ onto $\left(0 , \infty\right)$.

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:

${b}^{y} \ne 0 \text{ }$ for all Complex values of $y$

So for Complex valued logarithms:

${\log}_{b} x$ is only definable when $x \ne 0$

In fact, the function $y \to {b}^{y}$ is a continuous many to one function from $\mathbb{C}$ onto $\mathbb{C} \text{\} \left\{0\right\}$

If ${b}^{y} = x$ then ${b}^{y + \frac{2 n \pi i}{\ln b}} = x$ for any integer value of $n$

So for $x \ne 0$ we can pick a principal value of ${\log}_{b} x$ such that $I m \left({\log}_{b} x\right) \in \left(- \frac{\pi}{\ln b} , \frac{\pi}{\ln b}\right]$

With that convention, we would have:

${\log}_{b} x = \frac{\ln x}{\ln b} = \frac{1}{\ln b} \left(\ln \left\mid x \right\mid + A r g \left(x\right) i\right) = {\log}_{b} \left\mid x \right\mid + \frac{A r g \left(x\right)}{\ln b} i$

Whichever convention we use to pick the principal value of ${\log}_{b} x$, by definition we still have:

${b}^{{\log}_{b} x} = x$