If the point (10,2) lies on an exponential graph, then what point would have to lie on its inverse logarithmic graph?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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The point (10, 2) lies on the graph of an exponential function.

`f(x) = y = A*e^x`

`2 = A*e^10`

=> `A = 2/(e^10)`

The inverse logarithmic function is `f^-1(x)` such that

`(2/(e^10))*e^(f^-1(x)) = x`

take the natural log of both the sides

`f^-1(x)+ln(2/e^10) = ln x`

=> `f^-1(x) = ln x - ln 2 + 10`

`f^-1(2) = 10`

The inverse function for the function in reference is `f^-1(x) = ln x - ln 2 + 10`

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Let

`f(x)=A2^x`  is an exponential fuction .( Exponential functin may not defined in unique way)

Since point (10,2) lies on graph of f(x) therefore

`2=Axx2^10`

`A=2/2^10=2^(-9)`

`Thus`

`f(x)=2^(-9)xx2^x=2^(x-9)`

`` By def. of inverse of f ,we have

`f^(-1)(f(x))=x`

`f^(-1)(2^(x-9))=x`       (i)

Let

`2^(x-9)=y`

`log_2(y)=x-9`

`x=log_2(y)+9`       (ii)

Thus from (i) and (ii) ,we have

`f^(-1)(y)=log_2(y)+9`     (iii)

The point (2,10 ) lies on (iii).

`f^(-1)(2)=log_2(2)+9=1+9=10`

Thus if (10,2) lies on the graph of f(x) then (2,10) lies on `f^(-1)(x)`  .

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