Question

what is a solution to the given equation?
a 1
b 5
c 25
d 32

Explanation:

Step1: Substitute \(x = 1\)

$$
LATEXBLOCK0
$$

Step2: Substitute \(x = 8\)

$$
LATEXBLOCK1
$$

Step3: Substitute \(x = 25\)

$$
LATEXBLOCK2
$$

Step4: Substitute \(x = 32\)

$$
LATEXBLOCK3
$$

Wait, there is a mistake. The equation is \(4^{\frac{x}{2}}=1\). We know that \(a^{0}=1\) (\(a
eq0\)). So \(\frac{x}{2}=0\), \(x = 0\) is the solution. But if we assume the equation is \(4^{\frac{x}{2}}=1\) and check the options again (maybe mis - reading the original problem as \(4^{\frac{x}{2}} = 1\) and options are wrong - formatted). If the equation is \(4^{\frac{x}{2}}=1\), another way: \(4^{\frac{x}{2}}=(2^{2})^{\frac{x}{2}}=2^{x}\), and \(2^{x}=1\) when \(x = 0\). But if we assume the equation is \(4^{\frac{x}{2}}=1\) and check the options as per substitution (maybe the original equation is \(4^{\frac{x}{2}}=1\) and there is a typo in options). If we consider the general formula \(a^{b}=1\) ( \(a>0,a
eq1\)), then \(b = 0\).

Answer:

None of the options (A. 1, B. 8, C. 25, D. 32) are correct. If we assume the equation is \(4^{\frac{x}{2}}=1\), the solution is \(x = 0\). But if we force - check using the formula \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\):

• For \(x = 1\): \(4^{\frac{1}{2}}=\sqrt{4} = 2\)
• For \(x = 8\): \(4^{\frac{8}{2}}=4^{4}=256\)
• For \(x = 25\): \(4^{\frac{25}{2}}=(4^{\frac{1}{2}})^{25}=2^{25}\)
• For \(x = 32\): \(4^{\frac{32}{2}}=4^{16}=(2^{2})^{16}=2^{32}\)

If there is a mis - print and the equation is \(4^{\frac{x}{2}}=1\), the answer is not among the given options. If the equation is \(4^{\frac{x}{2}}=1\) and we consider the property \(a^{0}=1\) (\(a>0,a
eq1\)), \(x = 0\) is the solution.