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

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(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

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