Problems about Exponential and Logarithmic functions

READ THIS FIRST
Problems about Exponential and Logarithmic functions

READ THIS FIRST

These exercises are based on the theory treated on the page Exponential and Logarithmic functions

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about Exponential and Logarithmic functions

Level 1 problems

Show that x.e^-x +1 = 0 has exactly one root in [-1, -1/2]
1. Let y = x.e^-x +1 , then y' = e^-x.(1 - x)
  For x < 1 the function is increasing. So, in [-1, -1/2] there is at most one root.
2. For x = -1 , we have y < 0.
  For x = -1/2 , we have y > 0.
  So, there is a root in that interval [-1, -1/2] .

Solve:

 
   / ln(x) + ln(y²)=4
   \ (lnx)² - 3 ln(xy)= -5

with x > 0 and y > 0

 
   / ln(x) + ln(y²)=4
   \ (lnx)² - 3 ln(xy)= -5

<=>
   / ln(x) + 2.ln(y) = 4
   \ (lnx)² - 3 (ln(x)+ln(y)) = -5

  let ln(x) = u  ; ln(y) = v

<=>
   / u + 2v = 4
   \ u² - 3(u+v) = -5

<=>
   ....

<=>
    u = 2 ; v = 1  <=> x = e² ; y = e

or

   u = -1/2 ; v = 9/4  <=>  x = e^-1/2 ; y = e^9/4

Solve:

 
    5^{1/(2x - 1)} = 10^{(2x - 1)}

 
   Here is log the  log base 10

         1
<=>   ------- log(2) = 2x - 1
      2 x - 1


<=>  (2 x - 1)² = log(2)


<=>  2 x - 1 = sqrt(log(2))  or  2 x - 1 = - sqrt(log(2))

<=>  x  =  sqrt(log(2))/2 +1/2  or  x  =  - sqrt(log(2))/2 +1/2

Solve 15.3^x+1 - 243.5^x-2 = 0

 

        3.5. 3^x+1= 3⁵. 5^x-2

<=>        3^x-3 = 5^x-3

<=>        x - 3 = 0

<=>         x = 3

Find the equation of the inverse function of y = ln(2 - 1/e^x) .

 
     y  = ln(2 - 1/e^x)
<=>
     e^y = 2 - e^-x
<=>
     e^-x = 2 - e^y
<=>
     - x =  ln(2 - e^y)
<=>
      x = - ln(2 - e^y)

The equation of the inverse function is y = - ln(2 - e^x)

Simplify:
log_a y x u = ----------- log_a x y

 
               log_a y            log_a x
  ln(u) = ln( x        ) - ln ( y        )


        = log_a(y) ln(x) - log_a(x) ln(y)

           ln(y)           ln(x)
        = ------ ln(x) - ------- ln(y)
           ln(a)           ln(a)

        = 0

  u = 1

Level 2 problems

 

Solve              log_4/x(x²  - 6) = 2

 

Since 4/x is the base of a log function x is > 0.

The equation is then equivalent with

        (4/x)² = x²- 6

<=>     16 = x⁴- 6x²
                        let x² = t

<=>     16 = t²- 6t

<=>     t=-2  or t=8

<=>     x = sqrt(8)    (since x >0)

Calculate the slope of the tangent lines in point (2,0) of the graph of y = e^x.| x - 2 |

For x > 2 is y = e^x.( x - 2 ) and y' = e^x.( x - 1 )
The slope of the tangent line in point (2,0) is e².
For x < 2 is y = - e^x.( x - 2 ) and y' = - e^x.( x - 1 )
The slope of the tangent line in point (2,0) is - e².

 
Calculate the first and second derivative of

        y = x³. e^-x

 
        y' = 3x². e^-x- x³.e^-x

        y" = 6x.e^-x - 3x². e^-x - 3x². e^-x + x³.e^-x

        y" = 6x.e^-x - 6x². e^-x + x³.e^-x

 
Calculate the derivative of
        y = ln(tan(x/2))

 

                1               1             1
        y' = --------- .--------------------.---
              tan(x/2)   cos(x/2).cos(x/2)    2


                      1
           = -----------------------
                2 sin(x/2).cos(x/2)

                1
           =  ------
              sin(x)

 
Calculate the derivative of
        y = ln(tan(x/2 + pi/4))

 
Analogous as in the previous exercise, we find

        y' = 1/cos(x)

 
The function f(x) is given by

        (e^x-1)/x  for all x not 0
        1          for x = 0

Investigate if f(x) is continuous for x = 0

 

        f(x) is continuous for x = 0

<=>     lim f(x) = f(0)
         0

<=>     lim f(x) = 1
         0

                    e^x - 1
Now, lim f(x) = lim ------- =
      0          0    x

              e^x
        = lim ---- = 1
           0   1

Hence f(x) is continuous for x = 0

 
Find
             sin(x) + cos(x) - e^x
        lim ----------------------
         0      ln(1+ x²)

 
With l'Hospitals rule
              cos(x) -sin(x) - e^x
        = lim --------------------
           0    2x/(1 + x²)

                          cos(x) -sin(x) - e^x
        = lim (1 + x²) . --------------------
           0                    2x

                              cos(x) -sin(x) - e^x
        = lim (1 + x²) . lim --------------------
           0               0          2x

                - sin(x) - cos(x) - e^x
        1. lim -------------------------- = -1
            0           2

Given : f(x) = ln(e^-2 + e^x)
Prove that f(x) increases for all x.
What is the equation of the inverse function?
f'(x) = (e^-2 + e^x)^-1 . e^x > 0 for all x.
So, f(x) increases for all x.
The equation of the function f(x) is
```
 
   y =  ln(e^-2 + e^x)
<=>
   e^y = e^-2 + e^x
<=>
   e^x = e^-2 - e^y
<=>
   x = ln(e^-2 - e^y)
```
The equation of the inverse function is
```
 
   y =  ln(e^-2 - e^x)
```
Consider the function with equation f(x) =(x-m).e^m-x
Show that the maximum value of f(x) does not depend on the parameter m.
- First method
```
 
 f'(x) = e^m-x - (x-m) e^m-x

       = e^m-x (1-x+m)

 f(x) is maximum if x - m = 1

 The maximum is 1/e
```
- Second method
  A translation of the function to the left does not affect the maximum. We move the graph of f(x) m units to the left.
  The new equation is x.e^-x. The parameter no longer occurs.
  The maximum is independent of m.

Investigate the horizontal asymtote of f(x), for all m-values, as x tends to +infinity.

 
           e^{m x} + 1
   f(x) = ------------
              e^x

 
           e^{m x} + 1
     lim  ------------
    +infty     e^x


  =   lim  (e^{m x - x} + e^-x)
    +infty

  =   lim  e^(m-1)x
    +infty

 m = 1  => y = 1 is horizontal asymtote

 m < 1  => y = 0 is horizontal asymtote

 m > 1  =>  no horizontal asymtote

Find the domain of y = ln(2 - e^-x).
Find the vertical asymptote of the inverse function.
```
 
     x belongs to the domain of   y = ln(2 - e^-x).
<=>
     2 - e^-x > 0
<=>
     2 >  e^-x
<=>             Since the function ln(x) is rising
     ln(2) > - x
<=>
     x > -ln(2)
<=>
     x > ln(0.5)

The domain is ] ln(0.5) , + infty [
```
With a vertical asymptote of the inverse function corresponds a horizontal asymptote of the given function.
It is therefore unnecessary to find the inverse function.
We calculate the horizontal asymptotes of the given function.
In view of the domain there is only one horizontal asymptote.
```
 
   lim ln(2 - e^-x)  = ln(2)
  +infty
```
The only vertical asymptote of the inverse function is x = ln(2).

Level 3 problems

Solve

 
        x^(2ln(x)-1)   + e^(1/9) = (1 + e^(1/9)) x^(ln(x)-0.5)

 
Let y =   x^(ln(x)-0.5)

Then the given equation becomes

        y² +  e^(1/9) = (1 + e^(1/9)) y

This is a quadratic equation in y. The roots are

        y = 1  and y =  e^(1/9)
a)
        y = 1

<=>     x^(ln(x)-0.5)   = 1

<=>     (ln(x) - 0.5) = 0

<=>     ln(x) = 0.5

<=>     x = sqrt(e)

b)
        y = e^(1/9)

<=>     x^(ln(x)-0.5)   = e^(1/9)

Taking ln of both sides, we have

<=>     (ln(x) - 0.5).ln(x) = 1/9

<=>     ln(x).ln(x) - 0.5 ln(x) - 1/9 = 0

This is a quadratic equation in ln(x). The roots are

        ln(x) = 2/3  and ln(x) = -1/6

<=>     x = e^2/3  and  x = e^-1/6

 
Calculate  lim x^-x
             0

 

Instead of calculating the limit directly, we calculate

ln lim x^-x = lim ln(x^-x)
    0

                          -ln(x)
  = lim ( -x.ln(x)) = lim ------
     0                 0   1/x


With l'Hospitals rule we have

               1/x
        = lim ------- = lim x = 0
               1/(x²)
So,

                -x              -x
        ln lim x   = 0  => lim x   = 1
            0               0

 
Calculate the derivative of x^x

 

We know that

            ln(x^x)
    x^x = e         = e^{x ln(x)}


     d           d
So   --( x^x) = ---  e^{x ln(x)}
     dx         dx

                      d
        =  e^{x ln(x)} .  --( x.ln(x) )
                      dx



           x d
        = x .--(x.ln ( x ) )
             dx

           x
        = x .(ln(x) + 1)

You can generalize this method to calculate the derivative of

        g(x)
    f(x)

 
Find
               ln(ln(1+x⁴))
        lim ---------------
          0     ln(ln(1+x²))

 
With l'Hospitals rule

             ln(1+x²)   1+x²  4x³
     =  lim ----------.------.------
         0   ln(1+x⁴)   1+x⁴   2x


              2      2         2
             x ln(1+x )     1+x
     = 2 lim ----------.lim-----
          0        4     0     4
             ln(1+x )       1+x


             x².ln(1+x²)
     = 2 lim -------------
          0   ln(1+x⁴)

             x⁴.ln(1+x²)
     = 2 lim --------------
          0   x².ln(1+x⁴ )

by properties of logarithm

              ln(1+x²)^(1/x²)
     = 2 lim --------------------
          0    ln(1+x⁴)^(1/x⁴)

by properties of the number e
          ln e
     = 2 ------ = 2
          ln e

 
Investigate the function
                    e^x + 3 e^-x
        y =  ln(-----------------)
                     e^x + 1

 
The argument of ln is strictly positive. The domain of the function is R.

            e^x + 3 e^-x
y = 0 <=>  ------------- = 1
               e^x + 1


        <=>  e^x + 3 e^-x =  e^x + 1 <=>   3  =  e^x

        <=> x = ln 3


                  e^x - 6 - 3 e^-x
y' = ... = ----------------------------
               (e^x + 3 e^-x) (e^x + 1)


            e² ^x - 6 e^x - 3
        = --------------------------
          (e² ^x + 3) (e^x + 1)



let e^x= u then y' = 0 <=> u²- 6 u - 3 = 0

                       ___
        <=> u = 3 + 2 V 3
                          __
        <=> x = ln(3 + 2 V 3)

Investigation of the sign of y' tells that there is a minimum for
that x-value.

Since the domain is R, there are no vertical asymptotes

To find the horizontal asymptotes we calculate

                  e^x + 3 e^-x
        lim  ln(---------------)
      + infty        e^x  + 1



              x                  -x
             e                3 e
 = ln(lim --------  + lim  -------- ) = ln (1 + 0 ) = 0
             x                x
            e  + 1           e  + 1

y = 0 is a horizontal asymptote for x -> + infty

                  e^x + 3 e^-x
        lim  ln(---------------)
      - infty        e^x  + 1



              x                  -x
             e                3 e
 = ln(lim --------  + lim  -------- ) = ln (1 + infty) = + infty
             x                x
            e  + 1           e  + 1

There is not a horizontal asymptote for x -> - infty

To find the oblique asymptotes y = ax + b  we calculate

                          e^x + 3 e^-x
   a =       lim     ln(---------------) / x
            -infty            e^x  + 1


With l'Hospitals rule we find


               e² ^x - 6 e^x - 3
= lim     -------------------------- = -1
 -infty        (e² ^x + 3) (e^x + 1)



  b = lim  f(x) + x
    -infty

This limit is not so easy to calculate because of the presence of ln.
Therefore we calculate


         b       (f(x)+x)
        e = lim e
          -infty


                e^x + 3 e^-x
        = lim (--------------)  e^x
                 e^x + 1


                   e^2x+ 3
        = lim  (-----------) = 3
                 e^x + 1



It follows that b = ln 3.

The oblique asymptote is y = -x + ln 3

Solve next system for all real solutions

 
        e^x + e^-y² = 1                      (1)

        e^2x + sqrt( e^{- y²}) = 1             (2)

 
Let X = e^x   and Y = sqrt( e^{- y²})

Then X > 0 and Y > 0 and 2.ln(Y) =  - y² < 0

Thus,  0 < Y < 1

With this substitution the system becomes

        /
        |  X + Y² = 1        (3)
        |
        |  X² + Y = 1        (4)
        \

(3) - (4) gives

        X - X² + Y² - Y = 0

<=>     (X - Y) - (X² - Y²) = 0

<=>     (X - Y) ( 1 - X - Y ) = 0

<=>     X = Y or Y = 1 - X

First take Y = 1 - X .

        (4) becomes X² + 1 - X = 1

        This gives X = 0 or ( X = 1 and Y = 0 )

        Both cases give no solution for the original system.

Now take  X = Y

        (3) becomes  X² + X - 1 = 0

        The positive root is  X = (sqrt(5) - 1)/2

        So, X = Y = (sqrt(5) - 1)/2

        x =  ln ( (sqrt(5) - 1)/2 ) =  -0.481

and     e^{- y²} = Y² = (3 - sqrt(5))/2

<=>     - y² = ln ( (3 - sqrt(5))/2 )

<=>     y² = ln ( 2/(3 - sqrt(5)) )

<=>      y  = sqrt ( ln ( 2/(3 - sqrt(5)) ) ) = 0.981
     or  y  = - sqrt ( ln ( 2/(3 - sqrt(5)) ) ) = - 0.981

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