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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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Verified answer
Question :- Find the inverse of the function
[tex]\rm \: y = \dfrac{ {10}^{x} - {10}^{ - x} }{{10}^{x} + {10}^{ - x}} \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given function is
[tex]\rm \: y = \dfrac{ {10}^{x} - {10}^{ - x} }{{10}^{x} + {10}^{ - x}} \\ \\ [/tex]
can be rewritten as
[tex]\rm \: y = \dfrac{ {10}^{x} -\dfrac{1}{{10}^{x}} }{{10}^{x} + \dfrac{1}{{10}^{x}} } \\ \\ [/tex]
[tex]\rm \: y = \dfrac{\dfrac{{10}^{2x} - 1}{{10}^{x}} }{\dfrac{{10}^{2x} + 1}{{10}^{x}} } \\ \\ [/tex]
[tex]\rm \: y = \dfrac{{10}^{2x} - 1}{{10}^{2x} + 1} \\ \\ [/tex]
[tex]\rm \: y({10}^{2x} + 1) = {10}^{2x} - 1 \\ \\ [/tex]
[tex]\rm \: {10}^{2x}y +y = {10}^{2x} - 1 \\ \\ [/tex]
[tex]\rm \: {10}^{2x}y - {10}^{2x} = - y - 1 \\ \\ [/tex]
[tex]\rm \: {10}^{2x}(y - 1) = - (y + 1) \\ \\ [/tex]
[tex]\rm \: - {10}^{2x}(1 - y) = - (y + 1) \\ \\ [/tex]
[tex]\rm \: {10}^{2x}(1 - y) = y + 1 \\ \\ [/tex]
[tex]\rm \: {10}^{2x} = \dfrac{y + 1}{y - 1} \\ \\ [/tex]
We know,
[tex] \red{\boxed{ \sf{ \: {a}^{b} = c \: \rm\implies \:b = log_{a}(c) \: }}} \\ \\ [/tex]
So, using this result, we get
[tex]\rm \: 2x = log_{10}\bigg(\dfrac{y + 1}{1 - y} \bigg) \\ \\ [/tex]
[tex]\bf\implies \: x = \dfrac{1}{2} log_{10}\bigg(\dfrac{y + 1}{1 - y} \bigg) \\ \\ [/tex]
OR
[tex]\bf\implies \: {f}^{ - 1}(x) = \dfrac{1}{2} log_{10}\bigg(\dfrac{x + 1}{1 - x} \bigg) \\ \\ [/tex]