Question

pls solve





____full expain ​

Answer 1

Answer:

f^–1(X)=1/2log↓10(1+x/1–x)

Step-by-step explanation:

=10^2x–1/10^2x+1=y

=10^2x=1+y/1–y by componendo and dividendo

x=1/2log↓10(1+y/1-y)

f^-1(y)=1/2log↓10(1+y/1-y)

f^-1(x)=1/2log↓10(1+x/1-x)

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Answer 2

Question :- Find the inverse of the function

$\rm \: y = \dfrac{ {10}^{x} - {10}^{ - x} }{{10}^{x} + {10}^{ - x}}$

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y = \dfrac{ {10}^{x} - {10}^{ - x} }{{10}^{x} + {10}^{ - x}}$

can be rewritten as

$\rm \: y = \dfrac{ {10}^{x} -\dfrac{1}{{10}^{x}} }{{10}^{x} + \dfrac{1}{{10}^{x}} }$

$\rm \: y = \dfrac{\dfrac{{10}^{2x} - 1}{{10}^{x}} }{\dfrac{{10}^{2x} + 1}{{10}^{x}} }$

$\rm \: y = \dfrac{{10}^{2x} - 1}{{10}^{2x} + 1}$

$\rm \: y({10}^{2x} + 1) = {10}^{2x} - 1$

$\rm \: {10}^{2x}y +y = {10}^{2x} - 1$

$\rm \: {10}^{2x}y - {10}^{2x} = - y - 1$

$\rm \: {10}^{2x}(y - 1) = - (y + 1)$

$\rm \: - {10}^{2x}(1 - y) = - (y + 1)$

$\rm \: {10}^{2x}(1 - y) = y + 1$

$\rm \: {10}^{2x} = \dfrac{y + 1}{y - 1}$

We know,

$\red{\boxed{ \sf{ \: {a}^{b} = c \: \rm\implies \:b = log_{a}(c) \: }}}$

So, using this result, we get

$\rm \: 2x = log_{10}\bigg(\dfrac{y + 1}{1 - y} \bigg)$

$\bf\implies \: x = \dfrac{1}{2} log_{10}\bigg(\dfrac{y + 1}{1 - y} \bigg)$

OR

$\bf\implies \: {f}^{ - 1}(x) = \dfrac{1}{2} log_{10}\bigg(\dfrac{x + 1}{1 - x} \bigg)$