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How to find the inverse function of h(x)= 2^x ?

1 Answer

Apr 10, 2018

$h (x) = {log}_{2} (x)$ $h(x) = log_2(x)$

Explanation:

The find the inverse of an invertible function $y = f (x)$ $y = f(x)$ , swap $y$ $y$ and $x$ $x$ and solve for $y$ $y$ .

We have $h (x) = y = 2^{x}$ $h(x) = y = 2^x$ . Swapping $x$ $x$ and $y$ $y$ gives $x = 2^{y}$ $x = 2^y$ . We now wish to solve for $y$ $y$ .

Take the log with base 2 of both sides of the equation to free the $y$ $y$ variable.

$x = 2^{y}$ $x = 2^y$
${log}_{2} (x) = {log}_{2} (2^{y}) = y$ $log_2 (x) = log_2(2^y) = y$

Thus, $y = h (x) = {log}_{2} (x)$ $y = h(x) = log_2 (x)$ . See that we used the fact that ${log}_{n} (n^{x}) = x$ $log_n(n^x) = x$ .

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