Find the value of $\log_{21}{(x)}$ if $\log_{3}{(x)} = a$ and $\log_{7}{(x)} = b$

There are two logarithmic equations $\log_{3}{(x)} = a$ and $\log_{7}{(x)} = b$ are given in this logarithm problem and asked us to find the value of the term $\log_{21}{(x)}$.

Observe the bases of the logarithmic terms

The bases of the logarithmic terms in the given equations are $3$ and $7$. The product of them is equal to $21$ mathematically and it is base of the logarithmic term $\log_{21}{(x)}$ but it is not possible to multiply both equations directly to obtain the value of the log term $\log_{21}{(x)}$.

However, the value of the $\log_{21}{(x)}$ can be evaluated by expressing both equations in reciprocal form and then addition of them.

Express terms in reciprocal form

According to switch rule of logarithms, the bases and quantities in the log terms can be switched.

$(1) \,\,\,\,\,\,$ $\log_{x}{(3)} = \dfrac{1}{\log_{3}{(x)}} = \dfrac{1}{a}$

$(2) \,\,\,\,\,\,$ $\log_{x}{(7)} = \dfrac{1}{\log_{7}{(x)}} = \dfrac{1}{b}$

Addition of Logarithmic equations

Now, add the logarithmic equations to get their product by the product rule of the logarithms.

$\implies$ $\log_{x}{(3)} + \log_{x}{(7)}$ $\,=\,$ $\dfrac{1}{a} + \dfrac{1}{b}$

$\implies$ $\log_{x}{(3 \times 7)}$ $\,=\,$ $\dfrac{1}{a} + \dfrac{1}{b}$

$\implies$ $\log_{x}{(21)}$ $\,=\,$ $\dfrac{1}{a} + \dfrac{1}{b}$

$\implies$ $\log_{x}{(21)}$ $\,=\,$ $\dfrac{b+a}{ab}$

Switch the Logarithmic term

The value of $\log_{21}{(x)}$ has to be evaluated but the value of $\log_{x}{(21)}$ is evaluated in the previous step. If value of $\log_{x}{(21)}$ is expressed in reciprocal form, then the value of $\log_{21}{(x)}$ can be evaluated by the switch rule of logarithms.

$\implies$ $\log_{21}{(x)}$ $\,=\,$ $\dfrac{1}{\log_{21}{(x)}}$

$\implies$ $\log_{21}{(x)}$ $\,=\,$ $\dfrac{1}{\dfrac{b+a}{ab}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\log_{21}{(x)}$ $\,=\,$ $\dfrac{ab}{b+a}$