If x = log_{0.1} 0.001 and y = log_9 81, then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $x = \log_{0.1} 0.001$. Since $0.001 = (0.1)^3$, we have $x = \log_{0.1} (0.1)^3 = 3 \log_{0.1} 0.1 = 3(1) = 3$. Given $y = \log_9 81$. Sinc

Vedclass · Accepted Answer

$\sqrt{2} - 1$

Vedclass · Answer

(C) Given $x = \log_{0.1} 0.001$. Since $0.001 = (0.1)^3$, we have $x = \log_{0.1} (0.1)^3 = 3 \log_{0.1} 0.1 = 3(1) = 3$. Given $y = \log_9 81$. Since $81 = 9^2$, we have $y = \log_9 9^2 = 2 \log_9 9 = 2(1) = 2$. Now, we need to evaluate $\sqrt{x - 2\sqrt{y}}$. Substituting the values of $x$ and $y$, we get $\sqrt{3 - 2\sqrt{2}}$. We can write $3 - 2\sqrt{2}$ as $(\sqrt{2})^2 + (1)^2 - 2(\sqrt{2})(1) = (\sqrt{2} - 1)^2$. Therefore, $\sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$.

If $x = \log_{0.1} 0.001$ and $y = \log_9 81$, then $\sqrt{x - 2\sqrt{y}}$ is equal to

Similar Questions

If $n = (2020)!$,then the value of $\frac{1}{\log _{2} n} + \frac{1}{\log _{3} n} + \frac{1}{\log _{4} n} + \ldots + \frac{1}{\log _{2020} n}$ is equal to:

If $(\log _{5} x)(\log _{x} 3x)(\log _{3x} y) = \log _{x} x^{3}$, then $y$ equals

What is the number of real values of the parameter $k$ for which the equation $({\log _{16}}x)^2 - {\log _{16}}x + {\log _{16}}k = 0$ has exactly one solution,given that the coefficients are real?

Let $S$ be the sum of the digits of the number $15^2 \times 5^{18}$ in base $10$. Then,

If $A = \log _2 \log _2 \log _4 256 + 2 \log _{\sqrt{2}} 2$,then $A$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module