Question

The value of m for which $\left[\left\{\left(\frac{1}{7^2}\right)^{-2}\right\}^{-\frac{1}{3}}\right]^{\frac{1{4}} = 7^m$ is:}

• A. $-\frac{1}{3}$
• B. $\frac{1}{4}$
• C. -3
• D. 2

Hint:

When dealing with multiple nested exponents, work from the inside out. Always remember the fundamental rules of exponents: $(a^m)^n = a^{mn}$, $a^{-m} = \frac{1}{a^m}$, and $(a/b)^m = a^m/b^m$. Consistent application of these rules simplifies complex expressions.

Answer 1

The Correct Option is A

Step 1: Simplify the innermost expression using exponent rules.
\[ \left( \frac{1}{7^2} \right)^{-2 \times \left(-\frac{1}{3}\right) \times \frac{1}{4}} = 7^m \] Let's start with the innermost part: $\left(\frac{1}{7^2}\right)^{-2}$. Recall the exponent rule: $\left(\frac{1}{a^b}\right)^{-c} = (a^{-b})^{-c} = a^{bc}$.
Here, $a=7$, $b=2$, $c=2$.
So, $\left(\frac{1}{7^2}\right)^{-2} = (7^{-2})^{-2} = 7^{(-2) \times (-2)} = 7^4$.
Step 2: Substitute the simplified expression back and simplify the next layer.
Now the equation becomes: $\left[\left\{7^4\right\}^{-\frac{1}{3}}\right]^{\frac{1}{4}} = 7^m$. Next, simplify $\left(7^4\right)^{-\frac{1}{3}}$.
Recall the exponent rule: $(a^b)^c = a^{bc}$.
Here, $a=7$, $b=4$, $c=-\frac{1}{3}$.
So, $\left(7^4\right)^{-\frac{1}{3}} = 7^{4 \times (-\frac{1}{3})} = 7^{-\frac{4}{3}}$.
Step 3: Substitute this simplified expression back and simplify the outermost layer.
The equation is now: $\left[7^{-\frac{4}{3}}\right]^{\frac{1}{4}} = 7^m$. Finally, simplify $\left(7^{-\frac{4}{3}}\right)^{\frac{1}{4}}$.
Using the same rule $(a^b)^c = a^{bc}$:
Here, $a=7$, $b=-\frac{4}{3}$, $c=\frac{1}{4}$.
So, $\left(7^{-\frac{4}{3}}\right)^{\frac{1}{4}} = 7^{(-\frac{4}{3}) \times (\frac{1}{4})} = 7^{-\frac{4 \times 1}{3 \times 4}} = 7^{-\frac{4}{12}} = 7^{-\frac{1}{3}}$.
Step 4: Equate the exponents.
We have simplified the left side of the equation to $7^{-\frac{1}{3}}$. The original equation is $7^{-\frac{1}{3}} = 7^m$.
Since the bases are equal (both are 7), the exponents must also be equal. Therefore, $m = -\frac{1}{3}$.
(1) \quad \( -\frac{1}{3} \)