# 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.

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

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

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.

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

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

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

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