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How do you find the inverse of $y = ln (8 x + 1)$ $y=ln(8x + 1)$ ?

1 Answer

Dec 3, 2015

$x = \frac{e^{y} - 1}{8}$ $x=(e^y-1)/8$

Explanation:

If $y = ln (8 x + 1)$ $y=ln(8x+1)$
then (by definition)
$XXX e^{y} = 8 x + 1$ $color(White)("XXX")e^y=8x+1$

From this it follows that
$XXX e^{y} - 1 = 8 x$ $color(White)("XXX")e^y-1 = 8x$
and
$XXX x = \frac{e^{y} - 1}{8}$ $color(White)("XXX")x=(e^y-1)/8$

It is important to remember:
$XXX {log}_{b} a = c \Leftrightarrow b^{c} = a$ $color(White)("XXX")log_b a = c <=> b^c = a$

(and that $ln x = {log}_{e} x$ $ln x = log_e x$ )

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