Exponential Equations in Quadratic Form: Equation, Type, Questions

Exponential equations and quadratic equations are important components of algebra, each with distinct characteristics and applications. However, there are scenarios where exponential equations can be transformed into a quadratic form, allowing the use of techniques from quadratic equations to solve them. Exponential equations can be expressed in quadratic form, providing insights into their properties, solution methods, and applications.

Exponential Equations in Quadratic form

A polynomial equation in which the highest degree of a variable term is 2 is called a quadratic equation.

Standard form of quadratic equation is $a x^2+b x+c=0$

Where a, b, and c are constants (they may be real or imaginary) and called the coefficients of the equation and $a \neq 0$ (a is also called the leading coefficient).

Eg, $-5 x^2-3 x+2=0, x^2=0,(1+i) x^2-3 x+2 i=0$

As the degree of the quadratic polynomial is 2, so it always has 2 roots (number of real roots + number of imaginary roots = 2)

Roots of quadratic equation

The root of the quadratic equation is given by the formula:

$\begin{aligned} & x=\frac{-b \pm \sqrt{D}}{2 a} \\ & \text { or } \\ & x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\end{aligned}$

Where D is called the discriminant of the quadratic equation, given by $D=b^2-4 a c$,

An equation of the form a^x = b is known as an exponential equation, where

(i) $\mathrm{x} \in \phi$, if $\mathrm{b} \leq 0$
(ii) $x=\log _a b$, if $b>0, a \neq 0$
(iii) $\mathrm{x} \in \phi$, if $\mathrm{a}=1, \mathrm{~b} \neq 1$
(iv) $\mathrm{x} \in \mathrm{R}$, if $\mathrm{a}=1, \mathrm{~b}=1$ (since $1^{\mathrm{x}}=1 \Rightarrow 1=1, \mathrm{x} \in \mathrm{R}$ )

Some special cases of the exponential equation

1. Equation of the form $a^{f(x)}=1$, where $a>0$ and $a \neq 1$, then solve $f(x)=0$

For Example

The given equation is $7^{x^2+4 x+4}=1$
$\Rightarrow \mathrm{x}^2+4 \mathrm{x}+4=0 \quad\left[\because \mathrm{a}^0=1, \mathrm{a}\right.$ is constant $]$
$\Rightarrow(\mathrm{x}+2)(\mathrm{x}+2)=0 \Rightarrow \mathrm{x}=-2$

2. Equation of the form $f\left(a^x\right)=0$ then $f(t)=0$ where $t=a^x$

For example

The given equation is $4^x-3 \cdot 2^x-4=0$ equation is quadratic in $2^x$, so substitute $2^x=t$

$
\begin{aligned}
& \Rightarrow\left(2^{\mathrm{x}}\right)^2-3\left(2^{\mathrm{x}}\right)-4=0 \\
& \Rightarrow \mathrm{t}^2-3 \mathrm{t}-4=0 \\
& \Rightarrow(\mathrm{t}-4)(\mathrm{t}+1)=0 \\
& \Rightarrow \mathrm{t}=4, \mathrm{t}=-1
\end{aligned}
$

since, $\mathrm{t}=2^{\mathrm{x}}$

$
2^x=4 \Rightarrow 2^x=2^2 \Rightarrow x=2
$

and, $2^{\mathrm{x}}=-1$, No solution

Finally we get $x=2$

Summary

Exponential equations in quadratic form present an intriguing intersection of exponential and quadratic functions. By transforming exponential equations into quadratic form, we can apply familiar quadratic-solving techniques to find solutions. Understanding and mastering these transformations expand our ability to tackle a broader range of mathematical problems, enhancing both theoretical knowledge and practical problem-solving skills.

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Solved Examples Based on Exponential Equations in Quadratic form:

Example 1: If the sum of all the roots of the equation $\mathrm{e}^{2 x}-11 \mathrm{e}^x-45 \mathrm{e}^{-x}+\frac{81}{2}=0$ is $\log _{\mathrm{e}} \mathrm{p}$, then $\mathrm{p}$ is equal to______________.

1) 45

2) 17

3) 46

4) 16

Solution

The equation can be rewritten as

$2 e^{3 x}-22 e^{2 x}+81 e^x-90=0$

Let the roots are $\alpha, \beta, \gamma$

Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}_{,} \mathrm{e}^\alpha=\mathrm{t}_1, \mathrm{e}^\beta=\mathrm{t}_2, \mathrm{e}^\gamma=\mathrm{t}_3$

$2 t^3-22 t^2+81 t-90=0$

Product of roots $=\mathrm{t}, \mathrm{t}_2 \mathrm{t}_3=\mathrm{e}^\alpha \cdot \mathrm{e}^\beta \cdot \mathrm{e}^\gamma=\frac{90}{2}=45$

$\Rightarrow \mathrm{e}^{\alpha+\beta+\gamma}=45$

$\Rightarrow \alpha+\beta+\gamma=\log _e 45$

So $\mathrm{p}=45$

Hence, the answer is 45.

Example 2: The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking, it was found that, in the observations, 20 was misread as 5 . Then, the correct variance is equal to_______.

1) 17

2) 43

3) 81

4) 12

Solution

$\frac{\displaystyle\sum_{i=1}^{14} x i+5}{15}=8$
$\Rightarrow \displaystyle\sum_{i=1}^{14} x i=115 \Rightarrow \displaystyle\sum_{i=1}^{11} x i+20=123$
$\Rightarrow$ Real Mean $=\frac{\displaystyle\sum_{i=1}^{14} x i+20}{15}=\frac{13}{15}$
$\frac{\displaystyle\sum_{i=1}^{14} x i^2+5^2}{15}-(8)^2=9 \Rightarrow \displaystyle\sum_{i=1}^{14} x i^2=1070$
$\begin{aligned} \text { So Real variance } & =\frac{\displaystyle\sum_{i=1}^{14} \times i^2+20^2}{15}-(9)^2 \\ & =\frac{1070+400}{15}-(9)^2 \\ & =\frac{1470}{15}-81\end{aligned}$
$=98-81=17$

Hence, the answer is 17.

Example 3: The number of real roots of the equation $e^{6 x}-e^{4 x}-2 e^{3 x}-12 e^{2 x}+e^x+1=0$ is

1) 2

2) 4

3) 6

4) 1

Solution
$
\begin{aligned}
& e^{6 x}-e^{4 x}-2 e^{3 x}-12 e^{2 x}+e^x+1=0 \\
& \left(\left(c^{3 x}\right)^2-2 e^{3 x}+1\right)-c^x\left(e^{3 x}-1\right)-12 e^{2 r}=0 \\
& \Rightarrow\left(e^{3 x}-1\right)^2-e^x\left(e^{3 x}-1\right)-12\left(e^x\right)^2=0 \\
& \text { Let } e^{3 r}-1=p \text { and } e^2=q \\
& =p^2-p q-12 q^2=0 \\
& \Rightarrow p^2-4 p q+3 p q-12 q^2=0 \\
& B(p-4 q)(p+3 q)=0 \\
& E p=4 q \text { or } p=-3 q \\
& \Rightarrow e^{3 x}-1=4 e^x \text { or } e^{3 x}-1=-3 e^x \\
& \operatorname{Let} e^x=t(\therefore t>0) \\
& \Rightarrow t^3-4 t-1=0 \text { or } t^3+3 t-1=0 \\
& \operatorname{Let} f(t)=t^3-4 t-1 \& \operatorname{Let} g(t)=t^3+3 t-1 \\
& f^{\prime}(t)=3 t^2-4 \& g^{\prime}(t)=3 t^2+3
\end{aligned}
$

$f(t)$ has one positive root $\mathrm{t}=1 \& g(t)$ has one positive root (say $\mathrm{t}_1$ )
So 2 solutions

Example 4: The number of real roots of the equation $e^{4 r}-e^{3 r}-4 e^{2 r}-e^r+1=0$ is equal to $\qquad$.

1) 2

2) 0

3) 1

4) 3

Solution

$
e^{4 x}-e^{2 x}-4 e^{2 x}-e^x+1=0
$
Let $e^x=1$

$
\begin{aligned}
& t^1-t^4-4 t^2-t+1=0 \\
& \Rightarrow t^2-t-4-\frac{1}{t}+\frac{1}{t^2}=0 \\
& \Rightarrow\left(t^2+\frac{1}{t^2}\right)-\left(t+\frac{1}{t}\right)-4=0 \\
& \Rightarrow\left(t+\frac{1}{t}\right)^2-\left(t+\frac{1}{t}\right)-6=0
\end{aligned}
$
Let $t+\frac{1}{t}=t$

$
\begin{aligned}
& \Rightarrow u^2-u-6=0 \\
& \Rightarrow(u-3)(u+2)=0 \\
& \Rightarrow u=3,-2 \\
& \Rightarrow t+\frac{1}{t}=3 \quad\left(\text { As } t+\frac{1}{t}=c^x+\frac{1}{c^x}>0\right) \\
& \Rightarrow t^2-3 t+1=0 \\
& \Rightarrow t=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2} \\
& \Rightarrow e^x=\frac{3+\sqrt{5}}{2}, e^t=\frac{3-\sqrt{5}}{2} \\
& \Rightarrow x=\ln \left(\frac{3+\sqrt{5}}{2}\right) \cdot \ln \left(\frac{3-\sqrt{5}}{2}\right)
\end{aligned}
$
Hence, the answer is the option 1.

Example 5: The number of real roots of the equation $\mathrm{e}^{4 x}+2 \mathrm{e}^{3 x}-\mathrm{e}^x-6=\mathrm{n}$ is:

1)

2)

3)

4)

Solution

$f(x)=e^{4 x}+2 e^{3 x}-e^x-6$
$f^{\prime}(x)=4 e^{4 x}+6 e^{3 x}-e^x=0$$\begin{aligned} & e^x\left(4 e^{3 x}+6 e^{2 x}-1\right)=0 \\ & e^x>0 ; g(x)=4 e^{3 x}+6 e^{2 x}-1 \\ & a(-\infty)=-1: g(\infty)=\infty\end{aligned}$
and $g(x)$ is always increasing function
$\Rightarrow g(x)$ will have exactly one real root
$\Rightarrow f^{\prime}(x)$ will have exactly one real root which will be point of minima
Also $f(0)=-4$$\begin{aligned} & f(-\infty)=-6 \\ & f(\infty)=\infty\end{aligned}$

$\Rightarrow f(x)=0$ will have exactly 1 real root.