Download Inverse of An Exponential Function Since an exponential function, f

Inverse of An Exponential Function
Since an exponential function, f(x) = bx, is an 1-1 function, it has an inverse.
1
To find the inverse, we need to solve x b f ( x) .
The solution is defined to be f -1(x) = logbx.
• This is a logarithmic function with base b.
• b has the same restrictions as the exponential function, namely b > 0 and b ≠ 1.
• f-1(x) is the power b must be raised to in order to get x.
Examples
log232
log2
1
4
log55
log 1
log41
2
1
8
log 1 4
2
A Logarithmic Function
Form: f(x) = logbx where b > 0 and b ≠ 1.
Domain: x > 0.
Range: All real numbers.
Vertical Asymptote: x = 0.
Intercepts: (1, 0)
Increasing for x > 0 provided b > 1.
Decreasing for x > 0 provided 0 < b < 1.
No extreme values.
Graphs:
Special Logarithms
The common logarithm, logx, is the logarithm with base 10.
logx = log10x
The natural logarithm, lnx, is the logarithm with base e.
lnx = logex
Examples
log100
log0.1
log67
ln(e3)
ln9
ln0.245
log214
Transformations
All of the transformations apply.
Examples
1. f ( x) 3log 2 ( x 5) 2
2.
f ( x) 2ln( 3x) 1
Properties of Logarithms
1. logb1 = 0
Rationale: b0 = 1.
2. logbb = 1.
Rationale: b1 = b
3. logbbx = x for all real numbers.
Why is it true for all real numbers?
Rationale: Definition of inverses.
logb x
4. b
x for x > 0.
Why is it true only for x > 0?
Rationale: Definition of inverses.
Implication: Any exponential function can be converted to an exponential function in
another base.
• Examples
a. Since 2 = eln2, 2 x
b. Since 3 5log5 3 , 3x
eln 2
x
5log5 3
e(ln 2) x
x
5. logbx = logby if and only if x = y.
Rationale: 1-1 Property of Logarithms.
e0.693 x .
5(log5 3) x
50.6826 x .
Change of Base Formula
log a x
log b x
.
log a b
log x
and logb x
logb x
log b
Examples
log214
log350
ln x
are more practical.
ln b
log567
log20.75
log50.01
log 1 483
2
log 1 0.1
2
Additional Properties of Logarithms
Let x and y be positive numbers and n be any number.
1. logb(xy) = logbx + logby
Proof:
a. Let p = logbx and q = logby.
b. By the definition of logarithms, x = bp and y = bq.
c. Using the laws of exponents, xy = bpbq = bp + q.
d. Finally, logb(xy) = logb(bp + q) = p + q = logbx + logby.
2. logb(x/y) = logbx – logby
Proof: Similar to above.
3. logb(xn) = nlogbx
Proof: Similar to above.
Examples
6
log4x2y3
log5
2ln8 + 5ln(z – 4)
5log3x + 4log3y – 7log3z
x2 1
3
log 2
x2 y 4
2z3
½[log(x + 1) + 3log(x – 1)] + 6logx