Question

If ${\log }_{7}2=\lambda$, then find the value of ${\log }_{49}28$.
A. $(2\lambda +1)$
B. $(2\lambda +3)$
C. ${\displaystyle \frac{1}{2}}(2\lambda +1)$
D. $2(2\lambda +1)$

Answer 1

Hint: We first break the given term using factorisation. We take them to their simplest form. We use some logarithmic identities to find the value of ${\log }_{49}28$. We then put the value of ${\log }_{7}2=\lambda$ to find the final answer of the solution.

Complete step by step answer:
We have been given the value of one logarithm and we need to find another one using the previous one.
We are going to use some logarithmic identities ${\log }_{a^m}{b}^n={\displaystyle \frac{n}{m}}{\log }_{a}b,{\log }_a\left(xy\right)={\log }_{a}x+{\log }_{a}y,{\log }_{a}a=1$.
we now break the given term ${\log }_{49}28$ into its most simplistic form.
So, ${\log }_{49}28={\log }_{7^2}28={\displaystyle \frac{1}{2}}{\log }_{7}28$. The base of the logarithm is in its simplest form.
We try to break 28 into multiple forms. Prime factorisation of 28 is $28=(2^2\times 7)$.
${\log }_{49}28={\displaystyle \frac{1}{2}}{\log }_{7}28={\displaystyle \frac{1}{2}}{\log }_7(2^2\times 7)$.
Now we use identity ${\log }_a\left(xy\right)={\log }_{a}x+{\log }_{a}y$.
${\log }_{49}28={\displaystyle \frac{1}{2}}{\log }_7(2^2\times 7)={\displaystyle \frac{1}{2}}{\log }_7{2}^2+{\displaystyle \frac{1}{2}}{\log }_{7}7$.
We get the final answer by putting the value of ${\log }_{7}2=\lambda$
$\begin{array}{rl}& {\log }_{49}28\\ & ={\displaystyle \frac{2}{2}}{\log }_{7}2+{\displaystyle \frac{1}{2}}\left(1\right)\\ & ={\log }_{7}2+{\displaystyle \frac{1}{2}}\\ & =\lambda +{\displaystyle \frac{1}{2}}\end{array}$
We need to equate it with the given options which gives
${\log }_{49}28=\lambda +{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{2}}(2\lambda +1)$.

So, the correct answer is “Option C”.

Note: We use the logarithm values to break the powers of the log values. Logarithm identity ${\log }_{a^m}{b}^n={\displaystyle \frac{n}{m}}$ can be broken down into two parts for two different power values. Whenever we solve such questions, first we must look for clues from the question. We had to find the value of ${\log }_{49}28$ and we have been given ${\log }_{7}2=\lambda$ . So, we must be able to relate that 49 can be expressed as 7 to the power 2 and 28 is a multiple of 7, so again it can be expressed in terms of 7. Then we can easily apply the identities and simplify.