Question

4 what is the missing exponent in this equation? $2^{2} \times 3^{2} \times 4^{?} = 24^{2}$ a 8 b 1 c 6 d 2 5 which expression is equivalent to $3^{4} \cdot 4^{4}$? a $7^{8}$ b $7^{4}$ c $12^{8}$ d $12^{4}$

Explanation:

Question 4

Step1: Simplify left - hand side

First, we know that \(4 = 2^2\), so \(4^{2}=(2^{2})^{2}\). According to the power - of - a - power rule \((a^{m})^{n}=a^{mn}\), \((2^{2})^{2}=2^{2\times2}=2^{4}\).
Now, the left - hand side of the equation \(2^{2}\times3^{2}\times4^{2}\) can be rewritten as \(2^{2}\times3^{2}\times2^{4}\).
According to the product rule of exponents \(a^{m}\times a^{n}=a^{m + n}\), for the base 2 terms: \(2^{2}\times2^{4}=2^{2 + 4}=2^{6}\).
So the left - hand side becomes \(2^{6}\times3^{2}\). And we know that \(24=2^{3}\times3\), so \(24^{x}=(2^{3}\times3)^{x}\). According to the power - of - a - product rule \((ab)^{n}=a^{n}b^{n}\), \((2^{3}\times3)^{x}=2^{3x}\times3^{x}\).
We want to make the left - hand side equal to the right - hand side. Let's rewrite the left - hand side \(2^{2}\times3^{2}\times4^{2}\) in terms of 24.
First, \(2^{2}\times3^{2}\times4^{2}=(2\times3\times4)^{2}\) (because \(a^{n}\times b^{n}\times c^{n}=(a\times b\times c)^{n}\)).
Calculate \(2\times3\times4 = 24\), so \(2^{2}\times3^{2}\times4^{2}=24^{2}\).
The equation is \(24^{2}=24^{?}\), so the missing exponent is 2. Wait, there is a mistake above. Let's do it again.
We know that \(a^{n}\times b^{n}\times c^{n}=(a\times b\times c)^{n}\). So \(2^{2}\times3^{2}\times4^{2}=(2\times3\times4)^{2}\). Since \(2\times3\times4 = 24\), then \(2^{2}\times3^{2}\times4^{2}=24^{2}\).
The right - hand side is \(24^{?}\), so the missing exponent is 2.

Step2: Match with options

Looking at the options, option D is 2.

Step1: Recall the exponent rule

We know the rule \(a^{n}\times b^{n}=(a\times b)^{n}\) (this is the product of powers with the same exponent rule).
For the expression \(3^{4}\times4^{4}\), here \(a = 3\), \(b = 4\) and \(n = 4\).

Step2: Apply the rule

According to the rule \(a^{n}\times b^{n}=(a\times b)^{n}\), when \(a = 3\), \(b = 4\) and \(n = 4\), we have \(3^{4}\times4^{4}=(3\times4)^{4}\).
Calculate \(3\times4 = 12\), so \(3^{4}\times4^{4}=12^{4}\).

Answer:

D. 2

Question 5