What is the inverse of $y = {log}_{3} (4 x^{2} - 4)$ $y = log_3 (4x^2-4)$ ?

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Answer 1 · 2016-04-04T22:50:02

$y = \pm \sqrt{\frac{3^{x} + 4}{4}}$ $y=+-sqrt((3^x+4)/4)$

From the given equation $y = {log}_{3} (4 x^{2} - 4)$ $y=log_3 (4x^2-4)$

Interchange the variables, then solve for $x$ $x$

$x = {log}_{3} (4 y^{2} - 4)$ $x=log_3 (4y^2-4)$

$3^{x} = 3^{{log}_{3} (4 y^{2} - 4)}$ $3^x=3^(log_3 (4y^2-4))$

$y = \pm \sqrt{\frac{3^{x} + 4}{4}}$ $y=+-sqrt((3^x+4)/4)$

God bless ....I hope the explanation is useful.

What is the inverse of y = log_3 (4x^2-4) y=log3(4x2−4)y = log_3 (4x^2-4) ?