Question

If $t=-2$ then $lo{g}_4\frac{t^2}{4}-2lo{g}_{4}4t^4=$

• A. 2
• B. -4
• C. -6
• D. 0

Answer 1

The correct option is C -6

We have,

$t=-2$

Since,

${\log }_4\frac{t^2}{4}-2{\log }_{4}4t^4$

Therefore,

$={\log }_4\frac{{(-2)}^2}{4}-2{\log }_{4}4{(-2)}^4$

$={\log }_4\frac{4}{4}-2{\log }_{4}4\times 16$

$={\log }_{4}1-2{\log }_{4}64$

$=0-2{\log }_4{4}^3(\therefore {\log }_{a}1=0)$

$=-2\times 3{\log }_{4}4(\therefore {\log }_a{m}^n=n{\log }_{a}m)$

$=-6(\therefore {\log }_{a}a=1)$

Hence, this is the answer.