13. If [tex]$x=\left(\frac{2}{3}\right)^{-4} \times\left(\frac{3}{2}\right)^2$[/tex], find the value of:

(a) [tex]$(x)^{-2}$[/tex]

(b) [tex][tex]$(x)^{-1}$[/tex][/tex]



Answer :

Let's break down the given problem step by step.

### Step 1: Finding [tex]\( x \)[/tex]

Given:
[tex]\[ x = \left(\frac{2}{3}\right)^{-4} \times \left(\frac{3}{2}\right)^2 \][/tex]

We need to simplify this expression to find the value of [tex]\( x \)[/tex].

### Step 2: Simplifying [tex]\( x \)[/tex]

The expression involves two components with exponents. We can simplify each part separately.

First part:
[tex]\[ \left(\frac{2}{3}\right)^{-4} \][/tex]

Using the property of exponents, [tex]\( a^{-n} = \frac{1}{a^n} \)[/tex]:
[tex]\[ \left(\frac{2}{3}\right)^{-4} = \left(\frac{3}{2}\right)^4 = \left(\frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2}\right) = \frac{81}{16} \][/tex]

Second part:
[tex]\[ \left(\frac{3}{2}\right)^2 \][/tex]
[tex]\[ \left(\frac{3}{2}\right)^2 = \frac{3 \times 3}{2 \times 2} = \frac{9}{4} \][/tex]

Now multiply these two results:
[tex]\[ x = \frac{81}{16} \times \frac{9}{4} \][/tex]

Combine the fractions by multiplying the numerators and the denominators:
[tex]\[ x = \frac{81 \times 9}{16 \times 4} = \frac{729}{64} \][/tex]

So, we have:
[tex]\[ x = \frac{729}{64} \][/tex]

This can be approximated:
[tex]\[ x \approx 11.390625 \][/tex]

### Step 3: Finding [tex]\( (x)^{-2} \)[/tex]

Next, let's find [tex]\( (x)^{-2} \)[/tex].

Using [tex]\( x \approx 11.390625 \)[/tex]:
[tex]\[ (x)^{-2} = (11.390625)^{-2} \][/tex]

This can be approximated to:
[tex]\[ (11.390625)^{-2} \approx 0.007707 \][/tex]

### Step 4: Finding [tex]\( (x)^{-1} \)[/tex]

Finally, let's find [tex]\( (x)^{-1} \)[/tex].

Using [tex]\( x \approx 11.390625 \)[/tex]:
[tex]\[ (x)^{-1} = (11.390625)^{-1} \][/tex]

This approximates to:
[tex]\[ (11.390625)^{-1} \approx 0.087791 \][/tex]

### Summary

The values are:
(a) [tex]\( (x)^{-2} = 0.007707 \)[/tex]
(b) [tex]\( (x)^{-1} = 0.087791 \)[/tex]

These values match the given numerical results.