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The frictional force is a resisting force that occurs when two surfaces interact with each other. In our problem, the frictional force is computed between a rope and a wooden pole. The formula given is \( F = 14 \times 10^{0.70N} \), where \( F \) represents the frictional force in pounds and \( N \) represents the number of turns of the rope around the pole.
This relationship shows that as you increase the number of turns of the rope (\( N \)), the frictional force \( F \) also increases. Understanding this relationship helps in practical applications, such as ensuring that a rope tied around a pole stays secure.
In the step-by-step solution, we substitute different values of \( N \) to find how frictional force changes.

Exponentiation is a mathematical operation involving two numbers, the base and the exponent. It is written as \( a^b \), which means multiplying \( a \) by itself \( b \) times. In this exercise, exponentiation helps us understand how the frictional force increases with more turns of the rope.
Using the formula \( F = 14 \times 10^{0.70N} \), the exponent \( 0.70N \) determines the power to which 10 is raised. For example, when \( N = 0.5 \), we first calculate the exponent \( 0.70 \times 0.5 = 0.35 \) and then compute \( 10^{0.35} \).
This exponential increase is key to observing the impact of multiple turns on the frictional force.

The substitution method involves replacing a variable with a given value in an expression or equation. In our exercise, we substitute different values of \( N \) into the formula to calculate the corresponding frictional force.
For example, when \( N = 1 \), we substitute it into the equation \( F = 14 \times 10^{0.70 \times 1} \) and follow through with the calculations. This method allows us to evaluate specific scenarios and understand how each turn affects the frictional force.
It is a straightforward way to solve for unknowns and is widely used in algebra and physics problems.

Using a calculator is essential for complex calculations, especially those involving decimals and exponents.
In our exercise, after substituting the value of \( N \) and calculating the exponent, the next step involves using the calculator to find the value of \( 10^{exponent} \).
For instance, when \( N = 3 \), the exponent is 2.10: we use the calculator to find \( 10^{2.10} \), which leads to 126.491. Without a calculator, accurately evaluating such expressions would be very challenging and time-consuming.
Thus, practicing and familiarizing oneself with calculator functions is crucial for solving algebraic equations efficiently.