Logarithmic Equations and Inequalities

Interactive Feed

JEE Advanced 2013

If $3^{x} = 4^{x - 1}$ , then $x =$

\frac{2 l o g _{3} 2}{2 l o g _{3} 2 - 1}

(B)

\frac{2}{2 - l o g _{2} 3}

(C)

\frac{1}{1 - l o g _{4} 3}

(D)

\frac{2 l o g _{2} 3}{2 l o g _{2} 3 - 1}

JEE Advanced 2011

LEVELJEE Main

Let $(x_{0}, y_{0})$ be the solution of the following equations $(2 x)^{l n 2} = (3 y)^{l n 3}, 3^{l n x} = 2^{l n y}$ . Then $x_{0}$ is

The equation $x^{3/4 (l o g_{2} x)^{2} + l o g_{2} x - 5/4} = 2$ has

* Multiple Correct Options

(A)

at least one real solution

(B)

exactly three solutions

(C)

exactly one irrational solution

If $lo g_{0.3} (x - 1) < lo g_{0.09} (x - 1)$ , then $x$ lies in the interval

(A)

(2, \infty)

(B)

(1, 2)

(C)

(- 2, - 1)

The solution of equation $lo g_{7} lo g_{5} (x + 5 + x) = 0$ is .........

JEE Advanced 1986

LEVELBoard

The solution of the equation $lo g_{7} lo g_{5} (x + 5 + x) = 0$ is \dots.

JEE Advanced 1987

LEVELJEE Main

Solve for $x$ the following equation : $lo g_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - lo g_{(3 x + 7)} (4 x^{2} + 12 x + 9)$ .

JEE Advanced 1978

LEVELBoard

Solve the following equation for $x : 2 lo g_{x} a + lo g_{a x} a + 3 lo g_{a^{2} x} a = 0, a > 0, a \neq = 0$

JEE Main 2021 (31 Aug Shift 2)

LEVELJEE Main

The sum of the roots of the equation $x + 1 - 2 lo g_{2} (3 + 2^{x}) + 2 lo g_{4} (10 - 2^{- x}) = 0$ , is :

(A)

lo g_{2} 14

(B)

lo g_{2} 11

(C)

lo g_{2} 12

(D)

lo g_{2} 13

JEE Main 2021 (20 July Shift 2)

LEVELJEE Main

The number of solutions of the equation $lo g_{(x + 1)} (2 x^{2} + 7 x + 5) + lo g_{(2 x + 5)} (x + 1)^{2} - 4 = 0$ , $x > 0$ , is

JEE Main 2021 (26 February Shift 1)

LEVELJEE Main

The number of solutions of the equation $lo g_{4} (x - 1) = lo g_{2} (x - 3)$ is

JEE Main 2023 (11 Apr Shift 1)

LEVELJEE Main

The number of integral solution $x$ of $lo g_{(x + \frac{7}{2})} (\frac{x - 7}{2 x - 3})^{2} \geq 0$ is

JEE Main 2026 (23 January Shift 2)

LEVELJEE Main

The sum of all the real solutions of the equation $lo g_{(x + 3)} (6 x^{2} + 28 x + 30) = 5 - 2 lo g_{(6 x + 10)} (x^{2} + 6 x + 9)$ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 3x=4x−1, then x=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (x0​,y0​) be the solution of the following equations (2x)ln2=(3y)ln3,3lnx=2lny. Then x0​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation x3/4(log2​x)2+log2​x−5/4=2​ has

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If log0.3​(x−1)<log0.09​(x−1), then x lies in the interval

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of equation log7​log5​(x+5​+x​)=0 is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the equation log7​log5​(x+5​+x​)=0 is \dots.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solve for x the following equation : log(2x+3)​(6x2+23x+21)=4−log(3x+7)​(4x2+12x+9).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solve the following equation for x:2logx​a+logax​a+3loga2x​a=0,a>0,a=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the roots of the equation x+1−2log2​(3+2x)+2log4​(10−2−x)=0, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of the equation log(x+1)​(2x2+7x+5)+log(2x+5)​(x+1)2−4=0, x>0, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of the equation log4​(x−1)=log2​(x−3) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral solution x of log(x+27​)​(2x−3x−7​)2≥0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of all the real solutions of the equation log(x+3)​(6x2+28x+30)=5−2log(6x+10)​(x2+6x+9) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The product of all positive real values of x satisfying the equation x(16(log5​x)3−68log5​x)=5−16 is ______.

If $3^{x} = 4^{x - 1}$ , then $x =$

Let $(x_{0}, y_{0})$ be the solution of the following equations $(2 x)^{l n 2} = (3 y)^{l n 3}, 3^{l n x} = 2^{l n y}$ . Then $x_{0}$ is

The equation $x^{3/4 (l o g_{2} x)^{2} + l o g_{2} x - 5/4} = 2$ has

If $lo g_{0.3} (x - 1) < lo g_{0.09} (x - 1)$ , then $x$ lies in the interval

The solution of equation $lo g_{7} lo g_{5} (x + 5 + x) = 0$ is .........

The solution of the equation $lo g_{7} lo g_{5} (x + 5 + x) = 0$ is \dots.

Solve for $x$ the following equation : $lo g_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - lo g_{(3 x + 7)} (4 x^{2} + 12 x + 9)$ .

Solve the following equation for $x : 2 lo g_{x} a + lo g_{a x} a + 3 lo g_{a^{2} x} a = 0, a > 0, a \neq = 0$

The sum of the roots of the equation $x + 1 - 2 lo g_{2} (3 + 2^{x}) + 2 lo g_{4} (10 - 2^{- x}) = 0$ , is :

The number of solutions of the equation $lo g_{(x + 1)} (2 x^{2} + 7 x + 5) + lo g_{(2 x + 5)} (x + 1)^{2} - 4 = 0$ , $x > 0$ , is

The number of solutions of the equation $lo g_{4} (x - 1) = lo g_{2} (x - 3)$ is

The number of integral solution $x$ of $lo g_{(x + \frac{7}{2})} (\frac{x - 7}{2 x - 3})^{2} \geq 0$ is

The sum of all the real solutions of the equation $lo g_{(x + 3)} (6 x^{2} + 28 x + 30) = 5 - 2 lo g_{(6 x + 10)} (x^{2} + 6 x + 9)$ is equal to

The product of all positive real values of $x$ satisfying the equation $x^{(16 (l o g_{5} x)^{3} - 68 l o g_{5} x)} = 5^{- 16}$ is ______.