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**Hint:**When one term is raised to the power of another term, the function is called an exponential function, for example \[a = {x^y}\] . The inverse of the exponential functions are called logarithm functions, the inverse of the function given in the example is\[y = {\log _x}a\] that is a logarithm function. We can solve this using logarithmic rules, that is the Quotient rule of logarithm Product rule of logarithm Power rule of logarithm.

**Complete step by step solution:**

Given \[\log 6 + 2\log 5 + \log 4 - \log 3 - \log 2\]

If we apply logarithm rules directly in the problem it will be hard to simplify.

So we can see that in \[\log 6\], 6 can be written as multiple of 2 and 3

\[ = \log 6 = \log \left( {2 \times 3} \right)\]

Now applying the product rule, \[\log (x.y) = \log (x) + \log (y)\].

\[ = \log 6 = \log 2 + \log 3\]

Now the given problem becomes

\[ = \log 2 + \log 3 + 2\log 5 + \log 4 - \log 3 - \log 2\]

Cancelling \[\log 2\] and \[\log 3\] we have

\[ = 2\log 5 + \log 4\].

Again we can write 4 as multiple of 2 and 2

\[ = 2\log 5 + \log \left( {2 \times 2} \right)\]

Again applying the product rule, \[\log (x.y) = \log (x) + \log (y)\].

\[ = 2\log 5 + 2\log \left( 2 \right)\]

Now taking 2 common

\[ = 2\left( {\log 5 + \log 2} \right)\]

Now applying product rule \[\log (x.y) = \log (x) + \log (y)\]

\[ = 2\left( {\log \left( {5 \times 2} \right)} \right)\]

\[ = 2\left( {\log 10} \right)\]

We know \[\log 10 = 1\], then

\[ = 2\]

**Therefore, \[\log 6 + 2\log 5 + \log 4 - \log 3 - \log 2 = 2\].**

**Note:**

To solve this kind of problem we need to remember the laws of logarithms. Product rule of logarithm, that is the logarithm of the product is the sum of the logarithms of the factors. That is \[\log (x.y) = \log (x) + \log (y)\]. Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. That is \[\log \left( {\dfrac{x}{y}} \right) = \log x - \log y\]. Power rule of logarithm is that the logarithm of an exponential number is the exponent times the logarithm of the base. That is \[\log {x^a} = a\log x\]. These are the basic rules we use while solving a problem that involves logarithm function.

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