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

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.