Question

3^{log_{3}}^{7} * log_{4} \frac{1}{16}

Answer 1

$3 ^{log _{3}7 } *log _{4} \frac{1}{16}=7*(-2) = - 14$

Answer 2

Let's break down the expression step by step:

1. $3^{\log_{3}7}$: This can be simplified using the property that $a^{\log_{a}b} = b$. In this case, $a = 3$ and $b = 7$, so $3^{\log_{3}7} = 7$.

2. $log_{4}\frac{1}{16}$: We can rewrite $\frac{1}{16}$ as $4^{-2}$ since $4^{-2} = \frac{1}{4^2} = \frac{1}{16}$. Now, we have $log_{4}4^{-2}$, and using the property that $log_{a}a^{b} = b$, where $a = 4$ and $b = -2$, we get $log_{4}\frac{1}{16} = -2$.

Now, we can combine the simplified expressions:

$3^{\log_{3}7} \cdot log_{4}\frac{1}{16} = 7 \cdot (-2) = -14$

So, the value of the expression is $-14$.