Exponential Equations in Quadratic form - (Concept)

An equation of the form $\mathrm{a}^{\mathrm{x}}=\mathrm{b}$ is known as an exponential equation, where

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

Some special cases of the exponential equation:

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

For Example

The given equation is $7^{\mathrm{x}^2+4 \mathrm{x}+4}=1$ $\Rightarrow \mathrm{x}^2+4 \mathrm{x}+4=0\quad\left[\because\mathrm{a}^0=1,\mathrm{a}\right.$isconstant$]$$\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 $2^{\mathrm{x}}$, so substitute $2^{\mathrm{x}}=\mathrm{t}$

$
\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 \mathrm{x}=2
$

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