Equation Analysis and Simplification

The problem asks us to find the value of \(x\) in the given exponential equation: \[81 \times {\left( {\frac{{16}}{{25}}} \right)^{x + 2}} \div {\left( {\frac{3}{5}} \right)^{2x + 4}} = 144\] To solve this equation, our main strategy will be to express all numbers and fractions in terms of common bases (like prime numbers or simple fractions) and then use the rules of exponents to simplify the equation.

Converting Numbers to Common Bases

Let's convert the numerical terms in the equation to their base forms:

• The number \(81\) can be written as a power of \(3\): \(81 = 3 \times 3 \times 3 \times 3 = 3^4\).
• The fraction \(\frac{16}{25}\) can be written as a power of a fraction: \(\frac{16}{25} = \frac{4^2}{5^2} = {\left(\frac{4}{5}\right)^2}\).
• The number \(144\) can be written as a power of \(12\), and then further broken down: \(144 = 12^2 = (3 \times 4)^2 = 3^2 \times 4^2\). Since \(4 = 2^2\), we can write \(4^2 = (2^2)^2 = 2^4\). So, \(144 = 3^2 \times 2^4\).

Substituting and Simplifying the Exponential Equation

Now, substitute these simplified forms back into the original equation: \[3^4 \times {\left( {\left(\frac{4}{5}\right)^2} \right)^{x + 2}} \div {\left( {\frac{3}{5}} \right)^{2x + 4}} = 3^2 \times 2^4\]

Apply the exponent rule \((a^m)^n = a^{mn}\) to the second term: \[3^4 \times {\left( {\frac{4}{5}} \right)^{2(x + 2)}} \div {\left( {\frac{3}{5}} \right)^{2x + 4}} = 3^2 \times 2^4\] \[3^4 \times {\left( {\frac{4}{5}} \right)^{2x + 4}} \div {\left( {\frac{3}{5}} \right)^{2x + 4}} = 3^2 \times 2^4\]

Notice that the terms \(\left(\frac{4}{5}\right)^{2x + 4}\) and \(\left(\frac{3}{5}\right)^{2x + 4}\) have the same exponent. We can use the rule \(\frac{a^m}{b^m} = {\left(\frac{a}{b}\right)^m}\): \[3^4 \times {\left( {\frac{\frac{4}{5}}{\frac{3}{5}}} \right)^{2x + 4}} = 3^2 \times 2^4\] Simplify the fraction inside the parentheses: \[{\frac{\frac{4}{5}}{\frac{3}{5}}} = \frac{4}{5} \times \frac{5}{3} = \frac{4}{3}\] So the equation becomes: \[3^4 \times {\left( {\frac{4}{3}} \right)^{2x + 4}} = 3^2 \times 2^4\]

Now, distribute the exponent \((2x+4)\) to the numerator and denominator of \(\left(\frac{4}{3}\right)^{2x+4}\): \[3^4 \times \frac{4^{2x + 4}}{3^{2x + 4}} = 3^2 \times 2^4\]

Group the terms with base \(3\) using the rule \(\frac{a^m}{a^n} = a^{m-n}\): \[3^{4 - (2x + 4)} \times 4^{2x + 4} = 3^2 \times 2^4\] \[3^{4 - 2x - 4} \times 4^{2x + 4} = 3^2 \times 2^4\] \[3^{-2x} \times 4^{2x + 4} = 3^2 \times 2^4\]

Finally, express \(4\) as \(2^2\): \[3^{-2x} \times (2^2)^{2x + 4} = 3^2 \times 2^4\] Apply the rule \((a^m)^n = a^{mn}\) again: \[3^{-2x} \times 2^{2(2x + 4)} = 3^2 \times 2^4\] \[3^{-2x} \times 2^{4x + 8} = 3^2 \times 2^4\]

Solving for x by Equating Exponents

For the equation \(3^{-2x} \times 2^{4x + 8} = 3^2 \times 2^4\) to be true, the exponents of the corresponding bases on both sides of the equation must be equal.

• For base \(3\): The exponent on the left side is \(-2x\). The exponent on the right side is \(2\). Equating them: \[-2x = 2\] Divide both sides by \(-2\): \[x = \frac{2}{-2}\] \[x = -1\]
• For base \(2\): The exponent on the left side is \(4x + 8\). The exponent on the right side is \(4\). Equating them: \[4x + 8 = 4\] Subtract \(8\) from both sides: \[4x = 4 - 8\] \[4x = -4\] Divide both sides by \(4\): \[x = \frac{-4}{4}\] \[x = -1\]

Both equations yield the same value for \(x\), which is \(-1\).

Final Value of x

The value of \(x\) that satisfies the given equation is \(-1\).