Logarithms – JEE Main MathematicsPractice Questions & PYQs

Logarithmic equations often rely on recognizing algebraic patterns within the arguments and bases, allowing the expressions to be condensed into a manageable quadratic form. By using logarithmic properties to factorize and consolidate terms, the complex equation simplifies significantly, making it accessible through basic algebraic substitution.

Factorizing the quadratic expressions reveals a clear relationship between the logarithmic terms: $6x^{2}+28x+30 = 2(x+3)(3x+5)$ and $x^{2}+6x+9 = (x+3)^{2}$. Substituting these into the original equation $\log_{(x+3)}(6x^{2}+28x+30) = 5 - 2\log_{(6x+10)}(x^{2}+6x+9)$ allows us to expand the expression into $\log_{(x+3)}(x+3) + \log_{(x+3)}(6x+10) + 4\log_{(6x+10)}(x+3) = 5$. Since $\log_{(x+3)}(x+3)$ equals $1$, the equation reduces to $\log_{(x+3)}(6x+10) + 4\log_{(6x+10)}(x+3) = 4$.

Introducing the variable $t = \log_{(x+3)}(6x+10)$ transforms the expression into $t + \frac{4}{t} = 4$, which leads to the quadratic equation $t^{2} - 4t + 4 = 0$. Solving this yields $(t-2)^{2} = 0$, implying $t = 2$. Consequently, $\log_{(x+3)}(6x+10) = 2$ corresponds to $6x+10 = (x+3)^{2}$, or $x^{2}+6x+9 = 6x+10$, which simplifies to $x^{2} = 1$. Both $x = 1$ and $x = -1$ satisfy the domain constraints for the logarithms, as the bases remain positive and not equal to one. The sum of these real solutions is $1 + (-1) = 0$.

Infinite geometric series are evaluated using the sum formula $S = \frac{a}{1-r}$, where $a$ represents the first term and $r$ the common ratio. In the series $\alpha = \frac{1}{4} + \frac{1}{8} + \dots$, the first term is $\frac{1}{4}$ and the common ratio is $\frac{1}{2}$, resulting in $\alpha = \frac{1/4}{1-1/2} = \frac{1}{2}$. Similarly, the series $\beta = \frac{1}{3} + \frac{1}{9} + \dots$ has a first term of $\frac{1}{3}$ and a common ratio of $\frac{1}{3}$, which simplifies to $\beta = \frac{1/3}{1-1/3} = \frac{1}{2}$.

With both $\alpha$ and $\beta$ equal to $\frac{1}{2}$, the expression becomes $(0.2)^{\log_{\sqrt{5}}(1/2)} + (0.04)^{\log_{5}(1/2)}$. Expressing the bases as powers of 5, the first term is $(5^{-1})^{\log_{5^{1/2}}(1/2)}$. The exponent $\log_{5^{1/2}}(1/2)$ simplifies to $\frac{\log_5(1/2)}{1/2} = 2\log_5(1/2) = \log_5(1/4)$. Raising $5^{-1}$ to this power yields $5^{-\log_5(1/4)} = 5^{\log_5(4)} = 4$. The second term is $(5^{-2})^{\log_5(1/2)} = 5^{-2\log_5(1/2)} = 5^{\log_5(4)} = 4$. Adding these results produces the final value of 8.

By observing that the quadratic expression $2x^{2}+5x+3$ factors into $(2x+3)(x+1)$ and $x^{2}+2x+1$ is simply the square $(x+1)^{2}$, the given logarithmic equation can be rewritten as $\log_{(x+1)}((2x+3)(x+1)) = 4 - \log_{(2x+3)}(x+1)^{2}$. Using the product rule for the first term and the power rule for the second, the equation expands to $\log_{(x+1)}(2x+3) + \log_{(x+1)}(x+1) = 4 - 2\log_{(2x+3)}(x+1)$, which simplifies further to $\log_{(x+1)}(2x+3) + 1 = 4 - 2\log_{(2x+3)}(x+1)$ because $\log_{(x+1)}(x+1)$ equals 1. By substituting $t = \log_{(x+1)}(2x+3)$, we can express the second log term as $\frac{2}{t}$, turning the equation into $t + 1 = 4 - \frac{2}{t}$, or $t^{2} - 3t + 2 = 0$. Solving this quadratic results in $t=1$ or $t=2$. In the case where $t=1$, we have $\log_{(x+1)}(2x+3) = 1$, leading to $2x+3 = x+1$ and $x = -2$, which must be rejected as it violates the logarithmic domain constraint where the base must be positive and not equal to one. For the case where $t=2$, we have $\log_{(x+1)}(2x+3) = 2$, which implies $2x+3 = (x+1)^{2}$, or $x^{2} = 2$, yielding $x = \sqrt{2}$ and $x = -\sqrt{2}$. Since $x = -\sqrt{2}$ is also outside the valid domain for the logarithmic bases, the only viable solution is $x = \sqrt{2}$, and its square is 2.

Applying the natural logarithm to both sides of the equation $e^{5 (\log_e x)^2+3}=x^8$ allows for simplification by equating the exponents after taking the logarithm of the right side. This transformation yields the equation $5 (\log_e x)^2 + 3 = 8 \log_e x$. By substituting $t = \log_e x$, the expression converts into the standard quadratic form $5t^2 - 8t + 3 = 0$.

Factoring the quadratic equation as $5t^2 - 5t - 3t + 3 = 0$ leads to $(5t - 3)(t - 1) = 0$, which provides the two roots $t = 1$ and $t = \frac{3}{5}$. Converting these values back to $x$ gives the specific solutions $x = e^1$ and $x = e^{\frac{3}{5}}$. Multiplying these two results together requires adding their exponents, which produces $e^{1 + \frac{3}{5}}$, resulting in $e^{\frac{8}{5}}$.

The inequality
$\log_{\left(x+\frac{7}{2}\right)} \left(\frac{x-7}{2x-3}\right)^2 \ge 0$
is governed by specific constraints on the logarithm's base and argument. The base $x+\frac{7}{2}$ must be positive and not equal to one, while the argument must be non-zero, requiring $x > -\frac{7}{2}$, $x \neq -\frac{5}{2}$, $x \neq \frac{3}{2}$, and $x \neq 7$.

When the base is greater than one, occurring for $x > -\frac{5}{2}$, the argument must be at least one, satisfying the condition $\left(\frac{x-7}{2x-3}\right)^2 \ge 1$. Rearranging the squared terms as $(x-7)^2 \ge (2x-3)^2$ leads to $(2x-3)^2 - (x-7)^2 \le 0$, which simplifies to $(3x-10)(x+4) \le 0$, defining the interval $x \in [-4, \frac{10}{3}]$. The intersection of this interval with $x > -\frac{5}{2}$ restricts the valid range to $(-\frac{5}{2}, \frac{10}{3}]$, from which $\frac{3}{2}$ must be excluded.

When the base lies between zero and one, specifically for $-\frac{7}{2} < x < -\frac{5}{2}$, the argument must satisfy $0 < \left(\frac{x-7}{2x-3}\right)^2 \le 1$. The condition $\left(\frac{x-7}{2x-3}\right)^2 \le 1$ implies $x \in (-\infty , -4] \cup [\frac{10}{3}, \infty )$, which shares no common values with the valid interval for the base. Therefore, the only valid integers must fall within the range $(-\frac{5}{2}, \frac{10}{3}]$ while excluding $\frac{3}{2}$. The integers satisfying this condition are -2, -1, 0, 1, 2, and 3, resulting in 6 total solutions.

The core principle governing the existence of real roots in a quadratic equation is that its discriminant must be non-negative. To find the values for $\alpha$, we first apply logarithmic properties, specifically that the difference of logarithms with the same base is the logarithm of the quotient of their arguments, to rewrite the expression as $\log_2 \left( \frac{9^{2\alpha-4}+13}{\frac{5}{2} 3^{2\alpha-4}+1} \right) = 2$. By converting this to exponential form, the equation becomes $\frac{9^{2\alpha-4}+13}{\frac{5}{2} 3^{2\alpha-4}+1} = 4$. Substituting $y = 3^{2\alpha-4}$ leads to the quadratic equation $y^2 - 10y + 9 = 0$, which factors into $(y-9)(y-1) = 0$. Solving for $y$ gives $y=9$ and $y=1$, which correspond to $3^{2\alpha-4} = 3^2$ and $3^{2\alpha-4} = 3^0$. This yields $\alpha = 3$ and $\alpha = 2$.

With the set of values $S = \{2, 3\}$, we calculate the required sums to determine the coefficients of the quadratic equation. The sum of the elements in $S$ is $5$, and the sum of $(\alpha+1)^2$ for all $\alpha \in S$ equals $25$. Substituting these values into the given quadratic expression results in $x^2 - 50x + 25\beta = 0$. For this quadratic equation to have real roots, the discriminant, defined by $b^2 - 4ac$, must satisfy the inequality $50^2 - 4(1)(25\beta) \ge 0$. Simplifying this leads to $2500 - 100\beta \ge 0$, which indicates that $\beta \le 25$. Therefore, the maximum value of $\beta$ that allows for real roots is $25$.

Converting the logarithmic terms to a common base using the formula $\log_a b = \frac{\ln b}{\ln a}$ allows the equation $\log_{\cos x} \cot x + 4 \log_{\sin x} \tan x = 1$ to be rewritten as $\frac{\ln \cos x - \ln \sin x}{\ln \cos x} + 4 \frac{\ln \sin x - \ln \cos x}{\ln \sin x} = 1$. Expanding these fractions yields $1 - \frac{\ln \sin x}{\ln \cos x} + 4 - 4 \frac{\ln \cos x}{\ln \sin x} = 1$, which simplifies to $4 - \frac{\ln \sin x}{\ln \cos x} - 4 \frac{\ln \cos x}{\ln \sin x} = 0$. Multiplying the entire equation by the product $\ln \sin x \ln \cos x$ gives $4(\ln \sin x)(\ln \cos x) - (\ln \sin x)^2 - 4(\ln \cos x)^2 = 0$, which is the perfect square $-(\ln \sin x - 2 \ln \cos x)^2 = 0$.

This equality implies that $\ln \sin x = 2 \ln \cos x$, which is equivalent to $\sin x = \cos^2 x$. Substituting the trigonometric identity $\cos^2 x = 1 - \sin^2 x$ leads to the quadratic equation $\sin^2 x + \sin x - 1 = 0$. Using the quadratic formula to solve for $\sin x$ in the given domain, we obtain the positive root $\sin x = \frac{-1 + \sqrt{5}}{2}$. Comparing this expression with $\frac{\alpha + \sqrt{\beta}}{2}$ reveals that $\alpha = -1$ and $\beta = 5$, yielding a sum of $\alpha + \beta = 4$.

Applying logarithms to both sides of the exponential equations converts them into algebraic expressions, facilitating the relationship between the variables $a, b,$ and $c$. Taking the natural logarithm of $(2a)^{\log_e a} = (bc)^{\log_e b}$ yields $\log_e a (\log_e 2 + \log_e a) = \log_e b (\log_e b + \log_e c)$, while the second equation establishes the condition $\log_e 2 \cdot \log_e b = \log_e c \cdot \log_e a$. Solving these equations simultaneously reveals that the system is satisfied when either the product $ab=1$ or $bc=1$. Specifically, adopting the condition $bc=1$ in the second relationship forces $\log_e a = -\log_e 2$, which uniquely identifies $a=1/2$. Substituting these values into the expression $6a+5bc$ results in $6(1/2)+5(1)$, which simplifies to $8$.

Normalizing the bases to permit direct comparison of the logarithmic arguments is essential for solving logarithmic equations, provided that all resulting values satisfy the domain restrictions imposed by the original expressions. Converting the base 4 to base 2 relies on the property $\log_{4}(x-1) = \frac{1}{2}\log_{2}(x-1)$, which transforms the initial equation into $\frac{1}{2}\log_{2}(x-1) = \log_{2}(x-3)$. Multiplying both sides by 2 eliminates the fraction, resulting in $\log_{2}(x-1) = 2\log_{2}(x-3)$, which simplifies to $\log_{2}(x-1) = \log_{2}(x-3)^2$ using the power rule for logarithms.

Equating the arguments leads to the following algebraic steps:
$x-1 = x^2 - 6x + 9$
$x^2 - 7x + 10 = 0$
Factoring the quadratic produces $(x-5)(x-2) = 0$, yielding potential values of 2 and 5. Since the argument of the logarithm $\log_{2}(x-3)$ must be strictly positive, $x$ must be greater than 3, which necessitates rejecting the root 2. This leaves 5 as the only valid value, meaning there is exactly 1 solution to the equation.

The expression $2x^{2}+7x+5$ factors into $(2x+5)(x+1)$, allowing the first term to be written as $1 + \log_{x+1}(2x+5)$ by the product rule. Substituting this into the original equation yields $\log_{x+1}(2x+5) + \frac{2}{\log_{x+1}(2x+5)} = 3$, and setting $t = \log_{x+1}(2x+5)$ results in the quadratic $t^{2}-3t+2 = 0$, giving $t=1$ and $t=2$. For $t=1$, the condition $\log_{x+1}(2x+5) = 1$ implies $2x+5 = x+1$, giving $x=-4$, which is rejected. For $t=2$, $\log_{x+1}(2x+5) = 2$ leads to $2x+5 = (x+1)^{2}$, which simplifies to $x^{2}=4$, providing $x=2$ or $x=-2$. Only $x=2$ satisfies the domain $x>0$, confirming that exactly one solution exists.

By rearranging the logarithmic terms to one side and applying the property $\log_a M + \log_a N = \log_a (MN)$, the equation becomes
$\log_{1/2} |\sin x \cos x| = 2$
This logarithmic form is equivalent to the exponential equation
$|\sin x \cos x| = \frac{1}{4}$
Using the identity $\sin 2x = 2 \sin x \cos x$, we simplify this to
$|\sin 2x| = \frac{1}{2}$
Because the interval $[0, 2\pi]$ for $x$ corresponds to $2x \in [0, 4\pi]$, the equation $|\sin 2x| = 1/2$ produces 8 distinct solutions.