If {log _{12}}27 = a,then find the value of { | vedclass

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

Question

(C) Given that ${\log _{12}}27 = a$. Using the change of base formula,$\frac{{\log _3}27}{{\log _3}12} = a$. Since $27 = 3^3$ and $12 = 3 	imes 2^2$,

Vedclass · Accepted Answer

$4\frac{3-a}{3+a}$

Vedclass · Answer

(C) Given that ${\log _{12}}27 = a$. Using the change of base formula,$\frac{{\log _3}27}{{\log _3}12} = a$. Since $27 = 3^3$ and $12 = 3 	imes 2^2$,we have $\frac{3}{{\log _3}3 + {\log _3}2^2} = a$. This simplifies to $\frac{3}{1 + 2{\log _3}2} = a$. Rearranging gives $3 = a(1 + 2{\log _3}2)$,so $3 = a + 2a{\log _3}2$. Thus,$2a{\log _3}2 = 3 - a$,which means ${\log _3}2 = \frac{3-a}{2a}$. Now,we need to find ${\log _6}16 = \frac{{\log _3}16}{{\log _3}6} = \frac{{\log _3}2^4}{{\log _3}(2 	imes 3)} = \frac{4{\log _3}2}{{\log _3}2 + 1}$. Substituting ${\log _3}2 = \frac{3-a}{2a}$: ${\log _6}16 = \frac{4(\frac{3-a}{2a})}{\frac{3-a}{2a} + 1} = \frac{\frac{2(3-a)}{a}}{\frac{3-a+2a}{2a}} = \frac{2(3-a)}{a} 	imes \frac{2a}{3+a} = \frac{4(3-a)}{3+a}$. Therefore,the correct option is $C$.

If ${\log _{12}}27 = a$,then find the value of ${\log _6}16$.

Similar Questions

The set of real values of $x$ for which $\log_{0.2} \left( \frac{x + 2}{x} \right) \le 1$ is

If $\log _{(3x-1)}(x-2) = \log _{(9x^2-6x+1)}(2x^2-10x-2)$,then $x$ equals

If $\log_{10} 2 = 0.30103$ and $\log_{10} 3 = 0.47712$,then the number of digits in $3^{12} \times 2^8$ is:

If $x = \log_{5}(1000)$ and $y = \log_{7}(2058)$,then:

If $a^x = b^y = c^z = d^w$,then the value of $x\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{w}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module