In mathematics, if ƒ is a function from a set A to a set B, then an inverse function for ƒ is a function from B to A, Thus, if an input x into the function ƒ produces an output y, then input is y and inverse output is x. The inverse of a logarithmic function is an exponential function. If f is a function, then inverse function of f is f-1.
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loga(AB) = logaA + logaB
loga`(A/B)` = logaA - logaB
loga(Xr) = r logaX
Some other properties:
logaa = 1
loga1 = 0
loga`(1/x)` = - logax
logax = `(log x)/(log a)` = `(ln x) / (ln a)`
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Inverse logarithmic problem:
Determine the inverse function for the given expression ln(x - 5) + 3
we know, the function f(f-1(x)) = x and the base is e.
f(f-1(x)) = x --------> f(f-1(x)) = ln(f-1(x) - 5) + 3 = x
Subtracting by 3 on left and right side of the above equation.
ln(f-1(x) - 5) + 3 - 3 = x - 3
So, we get
ln(f-1(x) - 5) = x - 3
Now take exponential form on both side then we get,
`e^(x - 3) ` = f-1(x) - 5
Now add by 5 on both left and right side of the equation.
So we get, ` e^(x - 3)` + 5 = f-1(x) - 5 + 5
`e^(x - 3)` + 5 = f-1(x)
f-1(x) = `e^(x - 3)` + 5
The function f(x) is higher than 5 and the inverse function of f(x) is also higher than 5. inverse function of f(x) is f-1(x). if the line y = x is symmetry means, If the point (6, 3), graph the given function, f-1(3) = 6. That is point (3, 6) is graph of the function f-1(x) ,
`e^(x - 3)` + 5 = `e^(3 - 3)` + 5
= `e^(0)` + 5 we know e0 = 1
= 1 + 5
Answer: Inverse function of f(x) = 6.