Lesson Joke problem on solving exponential equation

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Joke problem on solving exponential equation


Problem 1

Solve this exponential equation and find  x:   3^[(x+1)/(x-5)] + 3^[(3x-9)/(x-5)] = 10/3.

Solution

        This equation looks scary.
        I will solve it informally.  It means that I will use
        my common sense in order for it lit my way in the darkness. 

Right side  10/3  is  10%2F3 = 31%2F3 = 3 + 1%2F3.


It tells me that one of the addend in the left side is 3, while the other is  1%2F3.


OK.   I try to have first addend equal to 3.   It leads me to this equation for exponent 

     %28x%2B1%29%2F%28x-5%29 = 1  -->  x+1 = x-5 ---> 1 = -5,   which is impossible,  so this way does not work.



OK.  I then try to have first addend equal to  1%2F3.  It leads me to this equation for exponent 

    %28x%2B1%29%2F%28x-5%29 = -1  -->  x+1 = -x + 5  -->  2x = 4  -->  x = 4/2 = 2.



With x= 2, the exponent in the second addend is  %283x-9%29%2F%28x-5%29 = %283%2A2-9%29%2F%282-5%29 = %28-3%29%2F%28-3%29 = 1.


It is exactly what I need.


So,   highlight%28highlight%28x=2%29%29  is the solution.    ANSWER

At this point, the solution is complete.

However, this solution does not guarantee that there is no another solution.

So, to check the answer, I used plotting tool DESMOS available online for free
www.desmos/calculator

You also can do it and repeat my steps.

Print equation of the function in the left side;  print another equation  y = 10/3  for the right side function.

The plot shows that the found solution  x= 2  is  highlight%28highlight%28UNIQUE%29%29.

Problem 2

If   x^(x¹⁶) = 16,  find   x²⁰.

Solution

        I will give absolutely rigorous mathematical solution,
        although it will contain element of guessing. 

Obviously, the value of x should be greater than 1.


One such value is easy to guess:  it is  x = 2^(1/4) = root%284%2C2%29 = 1.189207115,  approximately.


Indeed,  (2^(1/4))^16 = 2%5E4 = 16  and,  therefore,

    x^(x^16) = (2^(1/4))^16 = 2%5E4 = 16,


so this value of x,  x = 2^(1/4)  is the solution to the given equation  x^(x¹⁶) = 16.


From the other side hand,  this function, f(x) = x^(x¹⁶)  is a monotonically increasing function 
in the domain x > 1, so, there is no other solution to the given equation in this domain.



        Thus, for positive real numbers, the only solution 

            to the given equation  x^(x¹⁶) = 16  

        is  x = 2^(1/4) = root%284%2C2%29 = 1.189207115  (approximately).



CHECK.  1.189207115%5E16 = 16;  therefore,  x^(x¹⁶) = x^16 = 1.189207115^16 = 16,  which is correct.


Now,  x%5E20 = (2^(1/4))^20 = 2^5 = 32.


ANSWER.  Under given condition,  x%5E20 = 32.

Problem 3

If   2^a = 3,  3^b = 2,   find   1/(a+1) + 1/(b+1).

Solution 

If 2^a = 3  and  3^b = 2,  then  2^(ab) = (2^a)^b = 3^b = 2,  which implies  ab = 1.


Now,    1%2F%28a%2B1%29 + 1%2F%28b%2B1%29 = %28%28a%2B1%29+%2B+%28b%2B1%29%29%2F%28%28a%2B1%29%2A%28b%2B1%29%29 = 

      = %28a+%2B+1+%2B+b+%2B+1%29%2F%28ab+%2B+a+%2B+b+%2B+1%29 = %28a%2Bb%2B2%29%2F%281+%2B+a+%2B+b+%2B+1%29 = %28a%2Bb%2B2%29%2F%28a%2Bb%2B2%29 = 1.


ANSWER.  If  2^a = 3,  3^b = 2,  then  1%2F%28a%2B1%29 + 1%2F%28b%2B1%29 = 1.


My other lessons on exponential equations in this site are
    - Solving exponential equations
    - Solving advanced exponential equations
    - Solving exponential inequalities
    - Solving problems on population growth using logistic functions
    - Solving upper level exponential equations
    - OVERVIEW of lessons on solving exponential equations

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.


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