If log _{7} 2 = lambda, then the value of log | vedclass

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

Question

(A) Given that $\log _{7} 2 = \lambda$. We need to find the value of $\log _{49} (28)$. Using the change of base formula $\log _{a^n} b = \frac{1}{n}

Vedclass · Accepted Answer

$\frac{1}{2} (2 \lambda + 1)$

Vedclass · Answer

(A) Given that $\log _{7} 2 = \lambda$. We need to find the value of $\log _{49} (28)$. Using the change of base formula $\log _{a^n} b = \frac{1}{n} \log _{a} b$, we have: $\log _{49} (28) = \log _{7^2} (28) = \frac{1}{2} \log _{7} (28)$. Since $28 = 4 	imes 7 = 2^2 	imes 7$, we can write: $\frac{1}{2} \log _{7} (2^2 	imes 7) = \frac{1}{2} [\log _{7} (2^2) + \log _{7} 7]$. Using the property $\log _{a} (x^n) = n \log _{a} x$ and $\log _{a} a = 1$: $\frac{1}{2} [2 \log _{7} 2 + 1] = \frac{1}{2} (2 \lambda + 1)$.

If $\log _{7} 2 = \lambda$, then the value of $\log _{49} (28)$ is

Similar Questions

${\log _4}18$ is

If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$,then $x$ lies in which interval?

The expression $\log_{a} x$ is defined for $(a > 0, a \neq 1)$ when:

If $(x^2 \log _x 27) \cdot \log _9 x = x + 4$, then the value of $x$ is

The solution of $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module