Question

If t = -2, then find the value of \[{\log _4}\left( {\dfrac{{{t^2}}}{4}} \right) - 2{\log _4}\left( {4{t^4}} \right)\:\].
a) \[2\]
b) \[ - 4\]
c) \[ - 6\]
d) \[0\]

Answer 1

Hint: Here, in the question given to us, we would first of all, substitute the value of \[t\] in the given expression \[{\log _4}\left( {\dfrac{{{t^2}}}{4}} \right) - 2{\log _4}\left( {4{t^4}} \right)\:\]. Then we will further use the properties of logarithm to solve the expression \[{\log _4}\left( {\dfrac{{{t^2}}}{4}} \right) - 2{\log _4}\left( {4{t^4}} \right)\:\] for \[t = - 2\]. After using properties this equation turns into an easy addition and subtraction.
Formula used: Let \[a\] be the variable and base of the logarithm and be any \[y\] be any constant or variable that is power of variable\[a\]. Then we used the following properties of logarithm here.
\[
  {\log _a}1 = 0 \\
  {\log _a}a = 1 \\
  {\log _a}{a^y} = y \\
 \]

Complete step-by-step solution:
To solve the given question, we first put the value of \[t = - 2\] in the expression \[{\log _4}\left( {\dfrac{{{t^2}}}{4}} \right) - 2{\log _4}\left( {4{t^4}} \right)\:\]Hence,
\[
   \Rightarrow {\log _4}\left( {\dfrac{{{{( - 2)}^2}}}{4}} \right) - 2{\log _4}\left( {4{{( - 2)}^4}} \right)\;\, \\
   \Rightarrow {\log _4}\left( {\dfrac{4}{4}} \right) - 2{\log _4}\left( {4 \times 16} \right)\; \\
   \Rightarrow {\log _4}\left( 1 \right) - 2{\log _4}\left( {64} \right)\;\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} \to (1) \\
\]
Now we know that, \[{\log _a}1 = 0\] and \[{4^3} = 64\], then we use the given values in equation \[(1)\]
\[ \Rightarrow 0 - 2{\log _4}{\left( 4 \right)^3}\]
We now use another logarithmic property according to which
\[{\log _a}{a^y} = y\] and \[{\log _a}a = 1\].
Using this property we move ahead as
\[
   \Rightarrow - (2 \times 3){\log _4}\left( 4 \right) \\
   \Rightarrow - 6 \]
Thus from the above step we can say that the value of a given expression for a particular value \[t = - 2\] comes out to be c) \[ - 6\].

Note: This is to note here that the value of \[\log 1\] for any value of base is always zero. Value of log is never negative and its least value is zero. Logarithmic function is always a monotonously increasing function. Logarithmic function is the inverse of exponential function. When the log contains exponential \[e\] as base, then that log function is said to be a natural log.