Question

1. simplify. your answer should contain positive exponents only.

\\(\left(\frac{x^3}{x^5}\
ight)^{-2}\\)

1. solve for \\(x\\).

\\(3^{x - 7} = 27\\)

1. solve for \\(x\\).

\\(4^x \cdot 4^5 = 64^3\\)

Explanation:

Problem 1

Step1: Simplify the fraction inside the parentheses

First, simplify \(\frac{x^{3}}{x^{5}}\) using the rule of exponents \(a^m\div a^n=a^{m - n}\). So, \(\frac{x^{3}}{x^{5}}=x^{3-5}=x^{-2}\).

Step2: Apply the outer exponent

Now, we have \((x^{-2})^{-2}\). Using the power - of - a - power rule \((a^m)^n=a^{m\times n}\), we get \(x^{-2\times(-2)} = x^{4}\).

Step1: Rewrite 27 as a power of 3

We know that \(27 = 3^{3}\). So the equation \(3^{x - 7}=27\) can be rewritten as \(3^{x - 7}=3^{3}\).

Step2: Set the exponents equal

Since the bases are the same and the exponential function \(y = 3^{x}\) is one - to - one, we can set the exponents equal to each other: \(x-7 = 3\).

Step3: Solve for x

Add 7 to both sides of the equation: \(x=3 + 7=10\).

Step1: Simplify the left - hand side

Using the product rule of exponents \(a^m\times a^n=a^{m + n}\), for \(4^{x}\cdot4^{5}\), we get \(4^{x + 5}\).

Step2: Rewrite 64 as a power of 4

We know that \(64 = 4^{3}\), so \(64^{3}=(4^{3})^{3}\). Using the power - of - a - power rule \((a^m)^n=a^{m\times n}\), we have \((4^{3})^{3}=4^{3\times3}=4^{9}\).

Step3: Set the exponents equal

Now our equation is \(4^{x + 5}=4^{9}\). Since the bases are the same and the exponential function \(y = 4^{x}\) is one - to - one, we set the exponents equal: \(x + 5=9\).

Step4: Solve for x

Subtract 5 from both sides of the equation: \(x=9 - 5 = 4\).

Answer:

\(x^{4}\)

Problem 2