Answer :
Certainly! Let's evaluate the expression step by step:
Given the expression:
[tex]\[ 6 \left(1 - \frac{7}{16}\right)^{-\frac{1}{2} \times 2^{-1}} \][/tex]
### Step 1: Simplify the expression inside the parentheses
First, calculate the value inside the parentheses:
[tex]\[ 1 - \frac{7}{16} \][/tex]
Convert 1 into a fraction with the same denominator:
[tex]\[ 1 = \frac{16}{16} \][/tex]
So, we have:
[tex]\[ 1 - \frac{7}{16} = \frac{16}{16} - \frac{7}{16} = \frac{16 - 7}{16} = \frac{9}{16} \][/tex]
Thus, the expression becomes:
[tex]\[ 6 \left(\frac{9}{16}\right)^{-\frac{1}{2} \times 2^{-1}} \][/tex]
### Step 2: Simplify the exponent
Now we need to simplify the exponent [tex]\(-\frac{1}{2} \times 2^{-1}\)[/tex]:
Recall that [tex]\(2^{-1} = \frac{1}{2}\)[/tex]:
[tex]\[ -\frac{1}{2} \times \frac{1}{2} = -\frac{1 \times 1}{2 \times 2} = -\frac{1}{4} \][/tex]
So the exponent becomes [tex]\(-\frac{1}{4}\)[/tex], and the expression now is:
[tex]\[ 6 \left(\frac{9}{16}\right)^{-\frac{1}{4}} \][/tex]
### Step 3: Evaluate the power
Next, let's evaluate [tex]\(\left(\frac{9}{16}\right)^{-\frac{1}{4}}\)[/tex].
By definition of negative exponents:
[tex]\[ a^{-b} = \frac{1}{a^b} \][/tex]
So:
[tex]\[ \left(\frac{9}{16}\right)^{-\frac{1}{4}} = \frac{1}{\left(\frac{9}{16}\right)^{\frac{1}{4}}} \][/tex]
Now we need to find the fourth root ([tex]\( \frac{1}{4} \)[/tex]-th power) of [tex]\(\frac{9}{16}\)[/tex]:
[tex]\[ \left(\frac{9}{16}\right)^{\frac{1}{4}} = \left(\frac{3^2}{4^2}\right)^{\frac{1}{4}} = \left(\frac{3}{4}\right)^{\frac{1}{2}} \][/tex]
This implies:
[tex]\[ \frac{1}{\left(\frac{9}{16}\right)^{\frac{1}{4}}} = \frac{1}{\left(\frac{3}{4}\right)^{\frac{1}{2}}} \][/tex]
Next, evaluate the square root ( [tex]\( \frac{1}{2}\)[/tex]-th power) of [tex]\(\frac{3}{4}\)[/tex]:
[tex]\[ \left(\frac{3}{4}\right)^{\frac{1}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2} \][/tex]
So:
[tex]\[ \frac{1}{\left(\frac{3}{4}\right)^{\frac{1}{2}}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \][/tex]
### Step 4: Rationalize the denominator
To get a more simplified form we can rationalize the denominator:
[tex]\[ \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \][/tex]
Thus, the expression now becomes:
[tex]\[ 6 \times \frac{2\sqrt{3}}{3} \][/tex]
### Step 5: Simplify the final expression
Finally, simplify the multiplication:
[tex]\[ 6 \times \frac{2\sqrt{3}}{3} = \frac{6 \times 2\sqrt{3}}{3} = \frac{12\sqrt{3}}{3} = 4\sqrt{3} \][/tex]
By calculating the exact value, we get:
[tex]\[ 4\sqrt{3} \approx 6.928203230275509 \][/tex]
Therefore, the evaluated result of the given expression is approximately:
[tex]\[ 6.928203230275509 \][/tex]
Given the expression:
[tex]\[ 6 \left(1 - \frac{7}{16}\right)^{-\frac{1}{2} \times 2^{-1}} \][/tex]
### Step 1: Simplify the expression inside the parentheses
First, calculate the value inside the parentheses:
[tex]\[ 1 - \frac{7}{16} \][/tex]
Convert 1 into a fraction with the same denominator:
[tex]\[ 1 = \frac{16}{16} \][/tex]
So, we have:
[tex]\[ 1 - \frac{7}{16} = \frac{16}{16} - \frac{7}{16} = \frac{16 - 7}{16} = \frac{9}{16} \][/tex]
Thus, the expression becomes:
[tex]\[ 6 \left(\frac{9}{16}\right)^{-\frac{1}{2} \times 2^{-1}} \][/tex]
### Step 2: Simplify the exponent
Now we need to simplify the exponent [tex]\(-\frac{1}{2} \times 2^{-1}\)[/tex]:
Recall that [tex]\(2^{-1} = \frac{1}{2}\)[/tex]:
[tex]\[ -\frac{1}{2} \times \frac{1}{2} = -\frac{1 \times 1}{2 \times 2} = -\frac{1}{4} \][/tex]
So the exponent becomes [tex]\(-\frac{1}{4}\)[/tex], and the expression now is:
[tex]\[ 6 \left(\frac{9}{16}\right)^{-\frac{1}{4}} \][/tex]
### Step 3: Evaluate the power
Next, let's evaluate [tex]\(\left(\frac{9}{16}\right)^{-\frac{1}{4}}\)[/tex].
By definition of negative exponents:
[tex]\[ a^{-b} = \frac{1}{a^b} \][/tex]
So:
[tex]\[ \left(\frac{9}{16}\right)^{-\frac{1}{4}} = \frac{1}{\left(\frac{9}{16}\right)^{\frac{1}{4}}} \][/tex]
Now we need to find the fourth root ([tex]\( \frac{1}{4} \)[/tex]-th power) of [tex]\(\frac{9}{16}\)[/tex]:
[tex]\[ \left(\frac{9}{16}\right)^{\frac{1}{4}} = \left(\frac{3^2}{4^2}\right)^{\frac{1}{4}} = \left(\frac{3}{4}\right)^{\frac{1}{2}} \][/tex]
This implies:
[tex]\[ \frac{1}{\left(\frac{9}{16}\right)^{\frac{1}{4}}} = \frac{1}{\left(\frac{3}{4}\right)^{\frac{1}{2}}} \][/tex]
Next, evaluate the square root ( [tex]\( \frac{1}{2}\)[/tex]-th power) of [tex]\(\frac{3}{4}\)[/tex]:
[tex]\[ \left(\frac{3}{4}\right)^{\frac{1}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2} \][/tex]
So:
[tex]\[ \frac{1}{\left(\frac{3}{4}\right)^{\frac{1}{2}}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \][/tex]
### Step 4: Rationalize the denominator
To get a more simplified form we can rationalize the denominator:
[tex]\[ \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \][/tex]
Thus, the expression now becomes:
[tex]\[ 6 \times \frac{2\sqrt{3}}{3} \][/tex]
### Step 5: Simplify the final expression
Finally, simplify the multiplication:
[tex]\[ 6 \times \frac{2\sqrt{3}}{3} = \frac{6 \times 2\sqrt{3}}{3} = \frac{12\sqrt{3}}{3} = 4\sqrt{3} \][/tex]
By calculating the exact value, we get:
[tex]\[ 4\sqrt{3} \approx 6.928203230275509 \][/tex]
Therefore, the evaluated result of the given expression is approximately:
[tex]\[ 6.928203230275509 \][/tex]
