How do you find the inverse of $y = 6^{x} + 1$ $y=6^(x)+1$ ?

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Answer 1 · 2016-01-02T17:36:59

$y = {log}_{6} (x - 1)$ $y=log_6(x-1)$

Flip the $x$ $x$ and $y$ $y$ and solve for $y$ $y$ .

$x = 6^{y} + 1$ $x=6^y+1$

$x - 1 = 6^{y}$ $x-1=6^y$

To undo the exponentiation, take the logarithm with base $6$ $6$ of both sides.

${log}_{6} (x - 1) = {log}_{6} (6^{y})$ $log_6(x-1)=log_6(6^y)$

The logarithm will undo the exponent.

${log}_{6} (x - 1) = y$ $log_6(x-1)=y$

How do you find the inverse of y=6x+1y=6^(x)+1?