Inverse Functions and Logarithmic Functions - Desklib

This article covers topics on inverse functions and logarithmic functions. It explains how to find the inverse of a function, use composition to prove whether functions are inverses, and solve equations using logarithmic properties. Solved examples are provided for better understanding.

INVERSE FUNCTIONS
Inverse of the graph ( y=10x ¿
If a function f ( x ) is mapping x ¿ y , thenthe inverse function of f ( x ) maps y back ¿ x
f ( x )=10x ,interchan ging the variables x∧ y , we have :
x=10 y
Thus we solve for y .
Given that f ( x ) =g ( x ) , then ln f ( x ) =ln g ( x )
ln x=ln 10 y
Since log a xb =b∗log a x
ln 10 y= y∗ln10
ln x= y∗ln 10
y= ln x
ln10
Thus the inverse of y=10x is y= ln x
ln 10 .
Hence we graph y= ln x
ln 10
To graph the inverse function, we determine (x,y) points.

f−1 ( x ) if f ( x ) =−4 x +9
If a function f ( x ) is mapping x ¿ y , thenthe inverse function of f ( x ) maps y back ¿ x
y=−4 x +9
x=−4 y +9
−4 y +9= x
−4 y +9−9=x−9
−4 y=x−9
−4 y
−4 = x
−4 =−9
−4
y=−x−9
4
f−1 ( x ) =−x−9
4
Inverse of f ( x ) = ( x +1 ) 2−3
Simplifying f ( x ) = ( x+1 )2 −3
f ( x )=x2 +2 x−2
If a function f ( x ) is mapping x ¿ y , thenthe inverse function of f ( x ) maps y back ¿ x
x= y2 +2 y−2
y2 +2 y−2=x Subtracting x ¿ both sides , we have :
y2 +2 y−2−x=x −x
y2 +2 y−2=0
Using quadratic formula with a=1 , b=2, c=−2−x
y= √x +3−1
y=− √x +3−1
Thus Inverse of f ( x ) = ( x +1 ) 2−3 is : y= √ x+ 3−1∧ y=− √ x+ 3−1
Inverse of f ( x )=ln2 x

If a function f ( x ) is mapping x ¿ y , thenthe inverse function of f ( x ) maps y back ¿ x
y=ln 2 x
x=ln 2 y applying logarithmrule : a=log b ¿
x=ln ex
ln (ex ¿)=ln ( 2 y ) Given the logshave the same base : logb f ( x ) =log g b ( x ) ¿
Implying that f ( x ) =g ( x )
For ln ex=ln 2 y ,
we solve ex=2 y
y= ex
2
The inverse of Inverse of f ( x ) =ln 2 x is f ( x ) =e x
2
Inverse of the funct ion4.5 ( x−15 )4 +3 when x ≤15
Applyingbinomial theorem , we have :
( a+b )n=∑
i=0
n
(n
i )an−i bi Where a=x , b=−15
∑
i=0
n
(n
i )an−i bi=∑
i
4
(4
i )x4 −i (−15 )i
4.5 ( x−15 )4 +3=4.5 ∑
i
4
(4
i ) x4−i (−15 )i+3
f ( x )=4.5 x4− 67.5∗24
6 x3 + 1012.5∗24
4 x2−15187.5∗24
6 x+ 227815.5
If a function f ( x ) is mapping x ¿ y , thenthe inverse function of f ( x ) maps y back ¿ x
x=4.5 y 4− 67.5∗24
6 y3 + 1012.5∗24
4 y2− 15187.5∗24
6 y +227815.5
solving for y , we have 45 y4−2700 y3 +60750 y2−607500 y +2278155−x∗10=0

y= 1
3 (45−i2
1
4 √3 ( x−3 )
1
4 ), y= 1
3 (45+i 2
1
4 √3 ( x−3 )
1
4 )
Sinceboth roots are complex , Inverse of the function 4.5 ( x−15 ) 4 +3 when x ≤15 does not exist
Use composition ¿ prove whether the followin g functions are inverses :
If f ( g ( x ) )=g ( f ( x ) ) =x , then f ( x )∧g ( x ) are inverse functions
f ( x )=4 x +1
g ( x ) = x −1
4
f ( g ( x ) )=4 ( x−1
4 )+ 1= 4 x−4
4 =x −1
Hence the functions are not inverses
Use composition ¿ prove whether the follwoing functions areinverses . If they are not , find the inverse of
f ( x)
If f ( g ( x ) )=g ( f ( x ) ) =x , then f ( x )∧g ( x ) are inverse functions
f ( x )= ( x +1 )3−4
g ( x )= ( x−4 )
1
3 −1
f ( g ( x ) ) = [ ( x−4 )
1
3 −1+1 ]
3
−4= [ ( x−4 )
1
3 ]
3
−4
¿ x−4+4=x
Thus f ( g ( x ) )=g ( f ( x ) ) =x hence the functions areinverses
Logarithmic Functions
Question 1
20=4 ( e ) x
ln 20=ln( 4 ( e ) x ¿)=ln 4+ ln ex ¿
ln 20=ln 4 +x∗lne
ln 20=ln 4 +x

8=3x+ 4
ln 8=ln( 3¿¿ x+ 4)¿
44 x=200
ln 44 x=ln200
4 x∗ln 4=ln 200 x= ln200
4∗ln 4
Question 2
log5 20=x
5x=20
ln ( 100 ) =x
ex=100
2∗logx ( 20 )=4
logx 202=4
x4 =202
x4 =400
Question 3
Change of base formula: log b m= log a m
loga b
log3 5
log3 5= loge 5
loge 3 = ln 5
ln 3 =1.465
log6 22
log6 22= loge 22
loge 6 = ln22
ln6 =1.7251
log2 2.9

log2 2.9= loge 2.9
loge 2 = ln 2.9
ln 2 =1.536
Question 4
Product : logb xy=logb x +logb y
Quotient : logb
x
y =logb x −logb y
Power :log b x y= ylo gb x
Expand
a) log 24 ( 3 )
5
log 24 ( 3 )
5 =log 24 + log3−log 5
¿ 4 log 2−log3−log 5
b) log 25
9
log 25
9 =log 25−log 9
¿ log 52−log32
¿ 2 log 5−2 log 3
Contract
c) 6 log3 u+6 log3 v
¿ log3 u6 + log3 v6
¿ log3 u6 v6 =log3 ( uv )6
d) ln x−4 ln y
¿ ln x−ln y 4
¿ ln x
y4
Question 5
Convert to like bases:

a) 42 x +3=16
42 x +3=42
2 x+3=2
2 x=−1
x=−1
2
b) 32 a=3−a +9
2 a=−a+9
3 a=9
a=3
c) 53−2 x =1
53−2 x =50
3−2 x=0
−2 x=−3
x= 3
2
Use logs and log properties to solve:
a) 3b =17
b∗log 3=log 17
b= log17
log 3 =2.5789
b) 4 ( 5t ) −1=15
log 4 ( 5t ) =log ( 15+1 )
log 4+t∗log 5=log 16
t log 5=log 16−log 4
t= log 16−log 4
log5 =0.8614
c) ex−1 −5=5
ex−1=5+5=10
ex−1=10
( x−1 ) ln e=ln 10
x−1=ln 10
x=ln 10+1
x=3.3026
d) −6 (108 n+8 )−3=−23
−6 ( 108n +8 )=−23+3=−20
108 n +8=−20
−6 = 10
3

( 8 n+8 ) log10=log 10
3
8 n+8=
log 10
3
log 10
8 n=
log 10
3
log 10 −8=−7.47712
n=−7.47712
8 =−0.9346
e) a ¿ A=P ( 1+r ) t ,
where A isthe amount ∈the account , P isthe principleamount , r isthe interest rate∧¿
t is the period
b ¿ 2 p= p ( 1.02 ) t
2= ( 1.02 )t
log 2=log (1.02 )t
t= log 2
log 1.02 =35 months
¿ 2 years11 months
f) G=g ( 1.0045 )h
a ¿ Where G isthe numbe r of bacteria colonies after h hrs starting with g colonies
b ¿ 30000=8000 (1.0045 )h
log 30000=¿ log 8000+h log 1.0045 ¿
h= log30000−log 8000
1.0045
¿ 294.3839 hours
Exceeding
a) y=33+ 104 eert , when t=2 , y =120
120=33+104 e2 r
120−33=104 e2 r
87=104 e2 r
log 87=log 104+2 r
2 r=−0.077514086
r =−0.038757
Thus the rate of reduction is 0.038757 Degrees per minute
b) 80=33+104 e−0.038757 t
log 47=log 104±0.038757 t
−0.038757 t=−0.344935481
t=8.89995 ≈ 9

Thuswill take 9 minutes