Solve the problem.
-Given that $f ( x ) = 5 \ln 2 x$, find $f ^ { - 1 } ( x )$ and give the domain and range of $f ^ { - 1 } ( x )$.
A) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 1 } { 2 } \mathrm { e } ^ { x - 5 }$, domain $= ( - \infty , \infty )$, range $= ( 0 , \infty )$
B) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 1 } { 2 } \mathrm { e } ^ { \mathrm { x } } / 5$, domain $= ( - \infty , \infty )$, range $= ( 0 , \infty )$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 1 } { 2 } \mathrm { e } ^ { \mathrm { x } } / 5$, domain $= ( 0 , \infty )$, range $= ( - \infty , \infty )$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 1 } { 2 } \mathrm { e } ^ { \mathrm { x } - 5 }$, domain $= ( 0 , \infty )$, range $= ( - \infty , \infty )$

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The Answer of Solve the problem.
-Given that $f ( x ) = 5 \ln 2 x$, find $f ^ { - 1 } ( x )$ and give the domain and range of $f ^ { - 1 } ( x )$.
A) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 1 } { 2 } \mathrm { e } ^ { x - 5 }$, domain $= ( - \infty , \infty )$, range $= ( 0 , \infty )$
B) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 1 } { 2 } \mathrm { e } ^ { \mathrm { x } } / 5$, domain $= ( - \infty , \infty )$, range $= ( 0 , \infty )$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 1 } { 2 } \mathrm { e } ^ { \mathrm { x } } / 5$, domain $= ( 0 , \infty )$, range $= ( - \infty , \infty )$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 1 } { 2 } \mathrm { e } ^ { \mathrm { x } - 5 }$, domain $= ( 0 , \infty )$, range $= ( - \infty , \infty )$

Solve the Problem f(x)=5ln⁡2xf ( x ) = 5 \ln 2 xf(x)=5ln2x

Solve the Problem $f ( x ) = 5 \ln 2 x$