GATE ECE 2024 | GA | Question: 4

Question

​​​​For a real number $x>1$,
$$\frac{1}{\log _{2} x}+\frac{1}{\log _{3} x}+\frac{1}{\log _{4} x}=1$$ The value of $x$ is

• A. $4$
• B. $12$
• C. $24$
• D. $36$

Comments

option c

Answer 1

${\log 2 \over \log x} + { \log 3 \over \log x} + {\log 4 \over \log x} = 1 \to \log(x) = \log(4 \times 3 \times 2) \to x = 24.$

Comments

it is given that $x>1$ so no problem in taking $x$ as base.

$\dfrac{1}{log_2(x)} + \dfrac{1}{log_3(x)} + \dfrac{1}{log_4(x)} = 1$

$log_x(2) + log_x(3) + log_x(4) = 1$

$log_x(2 \times 3 \times 4) = 1$

$log_x(24) = 1$

$x^{1} = 24$

$x = 24$

Answer 2

Ans : c . 24

Answer 3

C

Comments

Can you give explanation for ever questions. Please!

Answer 4

option : C