Solving Exponential Equations (JEE , Olympiad Maths)-1

Here in this article on Solving Exponential Equations involving logarithmic functions, we will solve a good practice problem for clarifying the concepts on the subject matter for examination aspirants of CBSE, AIME, JEE, Olympiad Maths, IOQM, RMO aspirants and other Mathematics enthusiasts. Here we solve Exponents + Logarithms = A Brilliant Maths Challenge! 

Exponential Equation – Problem -1

Solve the equation for x and y when $(\frac{\ln x}{\ln y}{)}^{(x/y)}=(\frac{x}{y}{)}^{(\ln x/\ln y)}$ for (x+y)<100 and integer, x,y>1 and $x\ne y$

Step-by-Step Solution

The given problem is to find (x+y) such that its value is <100 and is integer. The exponential equation to be solved is $(\frac{\ln x}{\ln y}{)}^{(x/y)}=(\frac{x}{y}{)}^{(\ln x/\ln y)}$ , x,y>1 and $x\ne y$

For (x+y)<100 and to be integer means, x and y are also positive integers greater than 1.

Assumptions

Let $a=(\frac{\ln x}{\ln y})$ and $b=\frac{x}{y}$ .

Then $(\frac{\ln x}{\ln y}{)}^{(x/y)}=(\frac{x}{y}{)}^{(\ln x/\ln y)}\Rightarrow a^b=b^a$ .

The condition in the problem is $x\ne y$ but not $a\ne b$ .

Solution 1

If a=b condition is applied,

$\frac{\ln x}{\ln y}=\frac{x}{y}\Rightarrow \frac{\ln x}{x}=\frac{\ln y}{y}$

Let $f(t)=\frac{\ln t}{t}$ , then

The firs derivative of f(t) is given by,

$f^{\prime }(t)=\frac{\frac{1}{t}.t-\ln t.1}{t^2}=\frac{1-\ln t}{t^2}$

f(t) will be strictly increasing if $f^{\prime }(t)>0$, hence,

$\frac{1-\ln t}{t^2}>0\Rightarrow (1-\ln t)>0\Rightarrow \ln t<1\Rightarrow t<e$.

If t > 1 and for f(t) to be strictly increasing as f(t)>0, the value of t in the interval 0<t<e, t should be,

t=2 if t has to be integer.

The same rule if applied to $\frac{\ln x}{x}$ , then ,

x=2.

Then the value of y can be obtained from $\frac{\ln x}{x}=\frac{\ln y}{y}=\frac{\ln 2}{2}=\frac{2\ln 2}{2\times 2}=\frac{\ln 4}{4}$ .

If x= 2, y=4.

Hence,

Cross checking Solution

You can cross check the solution with equation if it is satisfied,

$(\frac{\ln x}{\ln y}{)}^{(x/y)}=(\frac{x}{y}{)}^{(\ln x/\ln y)}$

$(\frac{\ln x}{\ln y})=\frac{\ln 2}{\ln 4}=\frac{\ln 2}{2\ln 2}=\frac{1}{2}$

$\frac{x}{y}=\frac{2}{4}=\frac{1}{2}$ , hence,

LHS= $(\frac{\ln x}{\ln y}{)}^{(x/y)}$ = $(\frac{1}{2}{)}^{1/2}$

Similarly, RHS= $(\frac{x}{y}{)}^{(\ln x/\ln y)}$ = $(\frac{1}{2}{)}^{1/2}$ =LHS.

Hence solution is correct.

Solution 2

$\frac{\ln 16}{\ln 4}=2$and

$\frac{16}{4}=4=2^2$

Hence x=16 and y= 4 is another solution.

Hence as above there are two solution for the given problem of the given exponential equation. This answer for the exponential equation can be cross checked by substituting x and y in LHS and RHS.

The solution in video form can be accessible at link given for channel maths&science by iitian :

https://www.youtube.com/@mathsciencebyiitian

The image of the thumbnail for ease of identification of video is given as :

Conclusion

As above a problem on Exponential Equation involving the logarithmic functions is solved . We got two solutions for which(x+y)<100 and x and y being distinct , >1 and positive integers. the solution as x+y=6 ans x+y=20. If any other solutions exist can be checked for the given conditions.

In Algebra , the conditions are given and they will be applied in practice to solve the given problem with in the given bounds and inequalities.

If you have any clarifications on the subject solution please write to us at : https://mathsnscience.com/contact/

Best of luck to everybody.