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Use a calculator to evaluate the expression. Round your result to three decimal places.\((1.07)^{50}\)

Short Answer

Expert verified
The result is approximately 29.457 after rounding to three decimal places.

Step by step solution

01

Inputting Numbers

Input the base number \(1.07\) into the calculator.
02

Exponentiation

Apply the exponentiation operation with the exponent \(50\).
03

Rounding Off

Round off the result to three decimal places.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculator Usage
Using a calculator can make evaluating complex expressions really straightforward! Most scientific calculators come with a power or exponentiation button, often represented as
  • ^
  • EXP
  • or, occasionally, a button like xy
To evaluate expressions like \((1.07)^{50}\), start by typing the base number, which in this case is 1.07. Next, find and press the exponentiation button. After that, simply enter the exponent number, which is 50 here, and then hit the "equals" button, usually marked as =, to get your result.
Remember to double-check your entries while inputting numbers to prevent any typos or errors from affecting your final result. It only takes a moment, and it can save tons of time if you have to redo the calculations!
Rounding Numbers
Rounding numbers is a technique used to reduce the number of decimal places in a number. This is useful when more precision isn't necessary or when you want to simplify the number. In this exercise, you need to round the result to three decimal places.
Begin by identifying the number at the fourth decimal place. This digit will determine whether the third decimal place rounds up or stays the same. If the fourth digit is 5 or greater, add 1 to the third digit. If it's less than 5, leave the third digit as it is.
Here's an example: Suppose you calculated \((1.07)^{50}\) to be 120.381896. The number at the fourth decimal place is 8, which is greater than 5. So, round the number to 120.382 when rounding to three decimal places.
  • The third decimal place rounds up from 1 to 2
  • This makes your result more concise yet still accurate
Algebraic Expressions
Algebraic expressions are mathematical phrases that contain numbers, variables, and operations. They allow us to convey relationships and understand various mathematical concepts more deeply. In the given exercise, \((1.07)^{50}\), the expression illustrates how to deal with bases and exponents in algebra.
In such expressions, the number 1.07 is the base, and 50 is the exponent. The expression indicates that the base is being multiplied by itself 50 times. Understanding this helps you easily determine the operations required before reaching for your calculator.
Algebraic expressions like this one are fundamental in higher mathematics, appearing in areas like interest calculations, population growth, and more. If broken down properly, they reveal insights into how numbers interact with one another. Remember, the more familiar you become with expressions, the easier algebra will be!

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Most popular questions from this chapter

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(\left(1+\frac{0.065}{365}\right)^{365 t}=4\)

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln x-\ln (x+2)=3\)

The demand function for a special limited edition coin set is given by \(p=1000\left(1-\frac{5}{5+e^{-0.001 x}}\right)\) (a) Find the demand \(x\) for a price of \(p=\$ 139.50\). (b) Find the demand \(x\) for a price of \(p=\$ 99.99\). (c) Use a graphing utility to confirm graphically the results found in parts (a) and (b).

Domestic Demand The domestic demands \(D\) (in thousands of barrels) for refined oil products in the United States from 1995 to 2005 are shown in the table. (Source: U.S. Energy Information Administration)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Demand } \\ \hline 1995 & 6,469,625 \\ \hline 1996 & 6,701,094 \\ \hline 1997 & 6,796,300 \\ \hline 1998 & 6,904,705 \\ \hline 1999 & 7,124,435 \\ \hline 2000 & 7,210,566 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Demand } \\ \hline 2001 & 7,171,885 \\ \hline 2002 & 7,212,765 \\ \hline 2003 & 7,312,410 \\ \hline 2004 & 7,587,546 \\ \hline 2005 & 7,539,440 \\ \hline \end{array} $$(a) Use a spreadsheet software program to create a scatter plot of the data. Let \(t\) represent the year, with \(t=5\) corresponding to 1995 . (b) Use the regression feature of a spreadsheet software program to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use the regression feature of a spreadsheet software program to find a logarithmic model \((y=a+b \ln x)\) for the data. (d) Use a spreadsheet software program to graph the exponential model in base \(e\) and the logarithmic model with the scatter plot. (e) Use both models to predict domestic demands in 2008 , 2009, and \(2010 .\) Do both models give reasonable predictions? Explain.

Is it possible for a logarithmic equation to have more than one extraneous solution? Explain.

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