Quadratic Logarithmic Equations

Quadratic Logarithmic Equations

Komal MiglaniUpdated on 02 Jul 2025, 07:35 PM IST

Logarithmic equations are a fundamental aspect of algebra and are widely used in various scientific and engineering fields. Sometimes, logarithmic equations can be transformed into a quadratic form, enabling the use of techniques from quadratic equations to solve them. Logarithmic equations can be expressed in quadratic form, exploring their properties, solution methods, and applications.

This Story also Contains

  1. Logarithmic Equations in Quadratic form
  2. Summary
  3. Solved Examples Based on Logarithmic Equations in Quadratic form
Quadratic Logarithmic Equations
Quadratic Logarithmic Equations

Logarithmic 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 a 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 $D=b^2-4 a c$

Logarithmic Equations:

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 $\log _{\log _4 x} 4=2$
base of $\log$ is greater then 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 $\log _a f(x)=b \Rightarrow f(x)=a^b$

$
\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 $t=\log x$

$
\begin{aligned}
& \Rightarrow \frac{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}
$

Summary

Logarithmic equations in quadratic form present an intriguing intersection of logarithmic and quadratic functions. This approach is not only mathematically elegant but also highly practical, with applications in various fields such as mathematics, economics, physics, and engineering. 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.

Recommended Video Based on Quadratic Logarithmic Equations

Solved Examples Based on Logarithmic Equations in Quadratic form

Example 1: What is the solution of the inequation $\log _{x-3}\left(2\left(x^2-10 x+24\right)\right) \geq \log _{(x-3)}\left(x^2-9\right) ?$?

$\begin{aligned} & \text { 1) } x \in[4, \infty] \\ & \text { 2) } x \in(-\infty,-3) \cup[10+\sqrt{43, \infty} \\ & \text { 3) } x \in[10-\sqrt{43}, \infty] \\ & \text { 4) } x \in[10+\sqrt{43}, \infty]\end{aligned}$

Solution

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 $f(t)=0$

this inequation is equivalent to:

$
\begin{aligned}
& \left(2\left(x^2-10 x+24\right)\right) \geq\left(x^2-9\right) \\
& x^2-9>0 \\
& x-3>1
\end{aligned}
$

on solving these equation we get

$
x \in[10+\sqrt{43}, \infty)
$

Hence, the answer is the option 4.

Example 2: If for $x \in\left(0, \frac{\pi}{2}\right), \log _{10} \sin x+\log _{10} \cos x=-1$ and $\log _{10}(\sin x+\cos x)=\frac{1}{2}\left(\log _{10} n-1\right), n>0$ then the value of n is equal to:

1) 20

2) 16

3) 9

4) 12

Solution
$
\begin{aligned}
& \mathrm{x} \in\left(0, \frac{\pi}{2}\right) \\
& \log _{10} \sin x+\log _{10} \cos x=-1 \\
& \Rightarrow \quad \log _{10} \sin x \cdot \cos x=-1 \\
& \Rightarrow \quad \sin x \cdot \cos x=\frac{1}{10}
\end{aligned}
$

$\begin{aligned} & \log _{10}(\sin x+\cos x)=\frac{1}{2}\left(\log _{10} n-1\right) \\ & \Rightarrow \quad 2 \log _{10}(\sin x+\cos x)=\left(\log _{10} n-\log _{10}\right) \\ & \Rightarrow \quad(\sin x+\cos x)^2=10^{\left(\log _{10} \frac{n}{10}\right)} \\ & \Rightarrow \quad(\sin x+\cos x)^2=\frac{n}{10}\end{aligned}$

$\begin{aligned} \sin ^2 x+\cos ^2 x+2 \sin x \cdot \cos x & =\frac{n}{10} \\ \Rightarrow 1+\frac{1}{5}=\frac{n}{10} \quad \Rightarrow \quad n & =12\end{aligned}$

Hence, the answer is option 4.

Example 3: Let a complex number $z,|z| \neq 1$ satisfy $\log _{\frac{1}{\sqrt{2}}}\left(\frac{|z|+11}{(|z|-1)^2}\right) \leq 2$ Then, the largest value of |z| is equal to _________.

1) 7

2) 6

3) 5

4) 8

Solution

$\begin{array}{r}\log _{\frac{1}{\sqrt{2}}}\left(\frac{|z|+11}{(|z|-1)^2}\right) \leq 2 \\ \frac{|z|+11}{(|z|-1)^2} \geq \frac{1}{2} \\ 2|z|+22 \geq(|z|-1)^2 \\ 2|z|+22 \geq|z|^2+1-2|z| \\ |z|^2-4|z|-21 \leq 0 \\ (|z|-7)(|z|+3) \leq 0\end{array}$

$|z| \in[-3,7]$

$\therefore \quad$ Largest value of $|z|$ is 7

Hence, the answer is option 1.

Example 4: Solve the equation $2 \log _3 x+\log _3\left(x^2-3\right)=\log _3 0.5+5^{\log _5\left(\log _3 8\right)}$

1) $x=0$
2) $x=-2$
3) $x=2$

4) none of the above

Solution

Using properties of logarithm, this equation can be written as
$
\begin{aligned}
& \log _3 x^2+\log _3\left(x^2-3\right)=\log _3 0.5+\log _3 8 \\
& \log _3\left(x^2 \cdot\left(x^2-3\right)\right)=\log _3(0.5 * 8)
\end{aligned}
$

Now we have same base of $\log$ on both sides, so $\log$ can be removed from both sides

$
\begin{aligned}
& x^2\left(x^2-3\right)=0.5 \cdot 8 \\
& x^2\left(x^2-3\right)=4
\end{aligned}
$

$\begin{aligned} & x^4-3 x^2-4=0 \\ & \text { Let } x^2=t \\ & t^2-3 t-4=0 \\ & (t-4)(t+1)=0 \\ & t=4 \text { or } t=-1 \\ & x^2=4 \text { or } x^2=-1 \\ & x= \pm 2\end{aligned}$

Now check whether x= 2 and x = -2 lie in the domain of the original equation.

For x = -2, the first term in the equation is not defined. So it is rejected.

But for x = 2, all the terms are defined.

So x = 2 is the only answer.

Hence, the answer is the option 3.

Example 5: The number of solutions to the equation $\log _{(x+1)}\left(2 x^2+7 x+5\right)+\log _{(2 x+5)}(x+1)^2-4=0, x>0$ $\begin{aligned} & \log _{(x+1)}\left(2 x^2+7 x+5\right)+\log _{(2 x+5)}(x+1)^2-4=0 \\ \Rightarrow & \log _{(x+1)}((x+1)(2 x+5))+2 \log _{(2 x+5)}(x+1)-4=0\end{aligned}$ is:

1) 1

2) 2

3) 3

4) 4

Solution

$\begin{aligned} & \log _{(x+1)}\left(2 x^2+7 x+5\right)+\log _{(2 x+5)}(x+1)^2-4=0 \\ \Rightarrow & \log _{(x+1)}((x+1)(2 x+5))+2 \log _{(2 x+5)}(x+1)-4=0\end{aligned}$
$
\Rightarrow \log _{(x+1)}(x+1)+\log _{(x+1)}(2 x+5)+2 \log _{(2 x+5)}(x+1)-4=0
$

Let $\log _{(x+1)}(2 x+5)=t \Rightarrow \log _{(2 x+5)}(x+1)=\frac{1}{t}$
$
\Rightarrow \log _{(x+1)}(x+1)+\log _{(x+1)}(2 x+5)+2 \log _{(2 x+5)}(x+1)-4=0
$

Let $\log _{(x+1)}(2 x+5)=t \Rightarrow \log _{(2 x+5)}(x+1)=\frac{1}{t}$

$\begin{aligned} & \Rightarrow \quad 1+t+\frac{2}{t}-4=0 \\ & \Rightarrow \quad t^2-3 t+2=0 \\ & \Rightarrow \quad t=1 \quad \text { or } t=2\end{aligned}$

$\begin{aligned} & \Rightarrow \log _{(x+1)}(2 x+5)=1 \text { or } \log _{(x+1)}(2 x+5)=2 \\ & \Rightarrow 2 x+5=(x+1)^1 \text { or } 2 x+5=(x+1)^2 \\ & \Rightarrow x=-4 \quad \text { or } x^2=4\end{aligned}$

$\Rightarrow x=-4$ or $x=2$ or $x=-2$
Given $x>0 \Rightarrow x=2$

$x=2$ also lies in the domain of all the terms, so it is the answer.

Hence, the answer is the option (1).


Frequently Asked Questions (FAQs)

Q: How do you approach a quadratic logarithmic equation where the variable appears both inside and outside the logarithm?
A:
For such equations: 1) Try to isolate terms with the variable outside the logarithm.
Q: What is the importance of understanding the range of logarithmic functions in solving quadratic logarithmic equations?
A:
Understanding the range is important because: 1) It helps in identifying possible solutions. 2) It aids in recognizing impossible equations (e.g., log(x) = -5 has no real solutions). 3) It's crucial for graphical interpretations and solution verification. 4) It helps in understanding the behavior of the equation for different input values.
Q: How do you determine if a quadratic logarithmic equation will lead to a transcendental equation?
A:
A quadratic logarithmic equation may lead to a transcendental equation if: 1) After simplification, it can't be expressed as a polynomial equation. 2) It involves logarithms that can't be isolated or eliminated through algebraic means. 3) It requires numerical methods or special functions (like Lambert W) to solve. Recognizing this early can guide the choice of solving strategy.
Q: What role does the product rule of logarithms play in solving quadratic logarithmic equations?
A:
The product rule (log_a(xy) = log_a(x) + log_a(y)) is crucial because: 1) It allows splitting complex logarithmic expressions. 2) It can help in simplifying equations by separating quadratic terms. 3) It's often used in conjunction with the power rule to manipulate equations into solvable forms. 4) Understanding when and how to apply it is key to efficient problem-solving.
Q: How do you approach a system of quadratic logarithmic equations?
A:
To approach a system: 1) Try to simplify each equation individually using logarithmic properties. 2) Look for opportunities to substitute one equation into another. 3) Consider using elimination or substitution methods after transforming equations. 4) Be aware that the system may have multiple, one, or no solutions due to logarithmic constraints. 5) Check all solutions in both original equations.
Q: What is the significance of the base of logarithms in quadratic logarithmic equations?
A:
The base is significant because: 1) It determines which logarithmic properties can be applied. 2) Changing the base can simplify or complicate the equation. 3) Some bases (like e or 10) have special properties that can be leveraged. 4) Understanding the relationship between bases is crucial for solving equations with mixed bases.
Q: How do you handle quadratic logarithmic equations with absolute value terms?
A:
To handle absolute value terms: 1) Consider the equation in two cases: when the expression inside the absolute value is positive or negative. 2) Solve each case separately. 3) Combine solutions, ensuring they satisfy the original equation and domain restrictions. 4) Be aware that absolute value terms can introduce additional complexity and potential solutions.
Q: What is the role of asymptotes in the graphical representation of quadratic logarithmic equations?
A:
Asymptotes play a crucial role because: 1) Vertical asymptotes occur where the argument of the logarithm approaches zero, indicating domain restrictions. 2) They help visualize the behavior of the function as x approaches certain values. 3) Understanding asymptotes can provide insights into the number and nature of solutions. 4) They're essential for sketching accurate graphs of quadratic logarithmic functions.
Q: How do you determine the number of solutions a quadratic logarithmic equation will have?
A:
Determining the number of solutions involves: 1) Transforming the equation into a quadratic form. 2) Analyzing the discriminant of the resulting quadratic equation. 3) Considering domain restrictions from logarithms. 4) Checking for extraneous solutions. The final number of solutions may be less than or equal to the number suggested by the quadratic form due to logarithmic constraints.
Q: What is the importance of understanding function composition in quadratic logarithmic equations?
A:
Understanding function composition is crucial because quadratic logarithmic equations often involve the composition of logarithmic and quadratic functions. This understanding helps in: 1) Recognizing the structure of the equation. 2) Applying appropriate solving techniques. 3) Interpreting the meaning of solutions. 4) Visualizing the graphical representation of the equation.