${\log _7}{\log _7}\sqrt {7(\sqrt {7\sqrt 7 } )} = $
${\log _7}{\log _7}\sqrt {7(\sqrt {7\sqrt 7 } )} = $
- A
$3{\log _2}7$
- B
$1 - 3{\log _3}7$
- C
$1 - 3{\log _7}2$
- D
None of these
Similar Questions
The value of $\left(\left(\log _2 9\right)^2\right)^{\frac{1}{\log _2\left(\log _2 9\right)}} \times(\sqrt{7})^{\frac{1}{\log _4 7}}$ is. . . . . . .
The value of $\left(\left(\log _2 9\right)^2\right)^{\frac{1}{\log _2\left(\log _2 9\right)}} \times(\sqrt{7})^{\frac{1}{\log _4 7}}$ is. . . . . . .
- [IIT 2018]
If $\log x:\log y:\log z = (y - z)\,:\,(z - x):(x - y)$ then
If $\log x:\log y:\log z = (y - z)\,:\,(z - x):(x - y)$ then
If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$ then $x \ne 1$ lies in
If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$ then $x \ne 1$ lies in
The set of real values of $x$ for which ${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$ is
The set of real values of $x$ for which ${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$ is
Logarithm of $32\root 5 \of 4 $ to the base $2\sqrt 2 $ is
Logarithm of $32\root 5 \of 4 $ to the base $2\sqrt 2 $ is