Exponential Equations | Quadratic Functions | Algebra – Can you solve it?

In this article we will be solving exponential equations with embedded quadratic function. A quadratic function is used as an exponent in this exponential equation and is solved. It is a unique the of representation. We present two methods for solving this problem here. One by using pure algebra and another method by using calculus. You can choose any method to solve this problem.

Problem statement – Exponential Equation

Solve the below exponential equation for x.

$e^{(3x^2-5x+7)}=(3x^2-5x+7{)}^e$

Solution

Given,

$e^{(3x^2-5x+7)}=(3x^2-5x+7{)}^e$

We are required to solve for x. It is an exponential equation involving the Euler constant e and the quadratic function $(3x^2-5x+7)$.

It is not an ordinary exponential equation and not an ordinary quadratic equation. It should be dealt with separately.

Quadratic Equation inequality

Let the given quadratic equation $(3x^2-5x+7)$ be p. Then,

p= $(3x^2-5x+7)$.

Then the given exponential equation be written as,

e^p=p^e

But p = 3x²-5x+7 = $3(x-\frac{5}{6}{)}^2+\frac{59}{12}=\mathrm{is}\mathrm{always}\mathrm{positive}\Rightarrow$

$(3x^2-5x+7)>0\Rightarrow p>0$

Hence p is always positive.

Logarithmic property application

It is found that e^p=p^e and p is always positive.

Hence by taking natural logarithm on both sides , we get,

p=elnp

From above ,

$e=\frac{p}{\ln p}$

But we know that e is a constant and e> 0.

We also derived that p>0. Hence the denominator $\ln p\ne 0$.

From above,

$p\ne 1\mathrm{and}p>0$

Application of logarithmic inequality

It is known fact that for every, p>0,

$\ln p\le (p-1)$

Let $f(p)=p-e\ln p$ .

But if p=e, then,

$f(e)=e-e\ln e=0$

Hence the solution for $f(p)=p-e\ln p$ is,

$p=e$

Use of calculus to prove p=e

It is found from earlier discussion that $f(p)=p-e\ln p$.

By differentiating on both sides, we get,

$f^{‘}(p)=1-\frac{e}{p}$

From above, we get ,

$\begin{array}{c}{f}^{‘}(p)<0,\mathrm{for}0<p<e\\ f^{‘}(p)=0,\mathrm{for}p=e\\ f^{‘}(p)>0,\mathrm{for}p>e\end{array}$

From above,

$f(p)\mathrm{has}\mathrm{unique}\min \mathrm{imum}\mathrm{at}p=e$

Hence the unique solution for

$f(p)=0\mathrm{at}p=e$

Hence $p=e$.

Solving quadratic equation for x

Hence from $f(p)=0\mathrm{at}p=e$ , we get,

$(3x^2-5x+7)=e$

The above quadratic equation is written as,

$3x^2-5x+(7-e)=0$

By solving the quadratic equation, we get,

$x=\frac{5\pm \sqrt{25-12(7-e)}}{2\times 3}$

Hence from above,

$x=\frac{5\pm \sqrt{12e-59}}{6}$

As 12e-59<0, there are no real solutions.

Hence complex solutions for x are given as ,

$x=\frac{5\pm i\sqrt{59-12e}}{6}$

As above the problem of exponential equation has complex solution.

Conclusion

The given exponential equation involving the quadratic function doesn’t have real solution but has complex solution for x as,

$x=\frac{5\pm i\sqrt{59-12e}}{6}$

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A typical solution on complex plane is given as below.

In overall sense a combination of exponential equation coupled with quadratic function is solved both by using Pure algebra and also calculus as above.

Best of luck to everybody.