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What does the value of x if we solve $2 \log (3) + \log (x) = \log (36)$ ?

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answer is C.

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Detailed Solution

Given:

2 \log (3) + \log (x) = \log (36)

We know that the basic formulas of logarithms are:

log a + logb = \log (ab)

log a - logb = \log (\frac{a}{b})

And when no base has been given in the question, we assume that the base to be10. We also know that if

\log 10^{a} = c

then the value of a will be

(a) = 10^{c}

.
Now by applying the equation

loga + logb = \log (ab)

in the question, we can write

2 \log (3) + \log (x) = \log (36)

By using the law

\log a_{b}^{x} = x log a_{b}^{}

, we can write the above equation as

\Rightarrow \log (3^{2}) + \log (x) = \log (36)

\Rightarrow \log (9) + \log (x) = \log (36)

Now, by using the logarithmic rules on both the L.H.S and R.H.S of the above equation we get

\Rightarrow \log (9 x) = \log (36)

Now, taking antilogarithm on both sides, we get

\Rightarrow 9 x = 36

Now, to find the value of x we need to divide the right-hand side of the equation by the coefficient of x, that is 9. So we get,

\Rightarrow x = \frac{36}{9}

\Rightarrow x = 4

Now, if we put

x = 4

in the given equation, it satisfies both the L.H.S and R.H.S.
Thus, we can say that the value of

x = 4

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