Logarithmic Equations - (Concept)

Equation of the form $\log _{\mathrm{a}} \mathrm{f}(\mathrm{x})=\mathrm{b}(\mathrm{a}>0, \mathrm{a} \neq 1)$, is known as logarithmic equation. this is equivalent to the equation $\mathrm{f}(\mathrm{x})=\mathrm{a}^{\mathrm{b}}(\mathrm{f}(\mathrm{x})>0)$

Let us see one example to understand

Suppose given equation is $\log _{\log _4 x} 4=2$
the base of $\log$ is greater than 0 and not equal to 1 so, $\log _4 x>0$ and $\log _4 x \neq 1$
$x>1$ and $x \neq 4$
now, using $\log _{\mathrm{a}} \mathrm{f}(\mathrm{x})=\mathrm{b} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{a}^{\mathrm{b}}$

$
\begin{aligned}
& \Rightarrow 4=2^{\log _4 x} \Rightarrow 2^2=2^{\log _4 x} \\
& \Rightarrow 2=\log _4 x \Rightarrow x=4^2 \\
& x=16
\end{aligned}
$

If the given equation is in the form of $f\left(\log _a x\right)=0$, where $a>0$ and $a$ is not equal to 1 . In this case, put $\log _a x=t$ and solve $f(t)=0$.

And if the given equation is in the form of $f\left(\log _x A\right)=0$, where $A>0$. In this case, put $\log _x A=t$ and solve $f(t)=0$.

For example,

Suppose given equation is $\frac{(\log x)^2-4 \log x^2+16}{2-\log x}=0$ given equation can be written as after substituting $t=\log \mathrm{x}$

$
\begin{aligned}
& \Rightarrow \frac{\mathrm{t}^2-8 \mathrm{t}+16}{2-\mathrm{t}}=0 \\
& \Rightarrow \frac{(\mathrm{t}-4)(\mathrm{t}-4)}{(2-\mathrm{t})}=0 \\
& \Rightarrow \mathrm{t}=4 \\
& \mathrm{t}=\log \mathrm{x}=4 \\
& \because \log \mathrm{x}=\log _{10} \mathrm{x} \\
& \mathrm{x}=10^4
\end{aligned}
$