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Question

Question: How do you evaluate \({\log _6}2\) ?...

How do you evaluate log62{\log _6}2 ?

Explanation

Solution

In this question we have to use the identities of logarithm. Like first we need to convert its base into exponential using the formula logba=logealogeb{\log _b}a = \dfrac{{{{\log }_e}a}}{{{{\log }_e}b}}. Then we have to put the values of different logarithmic numbers.

Complete step by step answer:
In the above question it is given that log62{\log _6}2. We need to evaluate its value. We have to convert the base of the log into exponent so that we can evaluate its value. Otherwise, it would be difficult.
Therefore, we will use the formula logba=log10alog10b{\log _b}a = \dfrac{{{{\log }_{10}}a}}{{{{\log }_{10}}b}}to find the required value.
Here we have to substitute a=2andb=6a = 2\,\,and\,\,b = 6.
So, after substitution we get
log62=log102log106{\log _6}2 = \dfrac{{{{\log }_{10}}2}}{{{{\log }_{10}}6}}
Now we know that log102=0.693{\log _{10}}2 = 0.693 and log106=1.7917{\log _{10}}6 = 1.7917.
Now on substituting the above values, we have
log62=0.6931.7917{\log _6}2 = \dfrac{{0.693}}{{1.7917}}
log62=0.3868\Rightarrow {\log _6}2 = 0.3868
Therefore, the value of log62{\log _6}2 is 0.38680.3868.

Additional information: Logarithmic Functions have some of the properties that allow you to simplify the logarithms when the input is in the form of product, quotient or the value taken to the power. Some of the properties are as follows:
(1)\left( 1 \right) logb  MN = logb  M + logb  Nlo{g_{b\;}}MN{\text{ }} = {\text{ }}lo{g_b}\;M{\text{ }} + {\text{ }}lo{g_{b\;}}N
Multiply two numbers with the same base, then add the exponents.
(2)\left( 2 \right) logb  MN = logb  M  logb  Nlo{g_{b\;}}\dfrac{M}{N}{\text{ }} = {\text{ }}lo{g_b}\;M{\text{ }}-{\text{ }}lo{g_{b\;}}N
Divide two numbers with the same base, subtract the exponents.
(3)\left( 3 \right) Logb  Mp  = P logb  MLo{g_{b\;}}{M^p}\; = {\text{ }}P{\text{ }}lo{g_b}\;M
Raise an exponential expression to power and multiply the exponents.

Note: In this question we can also convert loge6=loge2+loge3{\log _e}6 = {\log _e}2 + {\log _e}3by using the property log(a×b)=loga+logb\log \left( {a \times b} \right) = \log a + \log b to make our question more easy. But I need to learn some values as we have used in this question. These are very used to values which are used frequently in questions of log functions.