If ${a^2} + 4{b^2} = 12ab,$ then $\log (a + 2b)$ is
If ${a^2} + 4{b^2} = 12ab,$ then $\log (a + 2b)$ is
- A
${1 \over 2}[\log a + \log b - \log 2]$
- B
$\log {a \over 2} + \log {b \over 2} + \log 2$
- C
${1 \over 2}[\log a + \log b + 4\log 2]$
- D
${1 \over 2}[\log a - \log b + 4\log 2]$
Similar Questions
The number of solution pairs $(x, y)$ of the simultaneous equations $\log _{1 / 3}(x+y)+\log _3(x-y)=2$ $2^{y^2}=512^{x+1}$ is
The number of solution pairs $(x, y)$ of the simultaneous equations $\log _{1 / 3}(x+y)+\log _3(x-y)=2$ $2^{y^2}=512^{x+1}$ is
- [KVPY 2017]
If ${\log _{\tan {{30}^ \circ }}}\left( {\frac{{2{{\left| z \right|}^2} + 2\left| z \right| - 3}}{{\left| z \right| + 1}}} \right)\, < \, - 2$ then
If ${\log _{\tan {{30}^ \circ }}}\left( {\frac{{2{{\left| z \right|}^2} + 2\left| z \right| - 3}}{{\left| z \right| + 1}}} \right)\, < \, - 2$ then
Let $\quad \sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c$, where $a, b, c \in Z$ and $e=\sum \limits_{n=0}^{\infty} \frac{1}{n!}$ Then $a^2-b+c$ is equal to $................$.
Let $\quad \sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c$, where $a, b, c \in Z$ and $e=\sum \limits_{n=0}^{\infty} \frac{1}{n!}$ Then $a^2-b+c$ is equal to $................$.
- [JEE MAIN 2023]
If $x = {\log _a}(bc),y = {\log _b}(ca),z = {\log _c}(ab),$then which of the following is equal to $1$
If $x = {\log _a}(bc),y = {\log _b}(ca),z = {\log _c}(ab),$then which of the following is equal to $1$
If ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x,$ then $x$ be
If ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x,$ then $x$ be