Find the value of minus $5$ raise to \[3\, \times \] minus $1$ raise to \[7\, \times 2\] raise to $2$?
Answer
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Hint: First we need to convert the given word problem into algebraic expression. That is an algebraic expression in mathematics is an expression which is made up of variables and constants along with algebraic operations. An expression is a group of terms. Here algebraic operations are addition, subtraction, division and multiplication etc. After obtaining the algebraic expression we can simplify.
Complete step by step solution:
We have minus 5 raise to \[3 \times \] minus 1 raise to \[7 \times 2\] raise to 2.
Now minus 5 raise to \[3\] means
\[ \Rightarrow {\left( { - 5} \right)^3}\]
Now minus 1 raise to \[7\], means
\[ \Rightarrow {\left( { - 1} \right)^7}\]
Now \[2\] raise to 2, means
\[ \Rightarrow {\left( 2 \right)^2}\].
Then minus 5 raise to \[3 \times \] minus 1 raise to \[7 \times 2\] raise to 2 becomes
\[ \Rightarrow {\left( { - 5} \right)^3} \times {\left( { - 1} \right)^7} \times {\left( 2 \right)^2}\]
Now we need to simplify this algebraic expression. We know the negative power of an odd number is negative only. That is \[ {\left( { - P} \right)^Q} = negative{\text{ }}value\], where P is any number and Q is an odd number.
\[ \Rightarrow \left( { - 125} \right) \times \left( { - 1} \right) \times \left( 4 \right)\]
\[ \Rightarrow \left( { - 125} \right) \times \left( { - 4} \right)\]
\[ \Rightarrow 500\].
Hence the required answer is 500.
Note:
Algebra helps in converting a mathematical statement into an equation. We know if we have ‘more’ or ‘sum’ in the given sentence we use the addition operation \[( + )\]. Similarly If we have ‘less’ or ‘difference’ we use subtraction \[( - )\]. If we have ‘quotient’ we use division operation \[( \div )\]. To define more generalized terms; we use algebra. It is a very vast branch of mathematics and is used in all the branches of mathematics like polynomial, linear equations, graphs, etc. and in daily life too.
We know that the product of two negative numbers results in a positive number. The product of a positive (negative) number and a negative (positive) number results in a negative number.
Complete step by step solution:
We have minus 5 raise to \[3 \times \] minus 1 raise to \[7 \times 2\] raise to 2.
Now minus 5 raise to \[3\] means
\[ \Rightarrow {\left( { - 5} \right)^3}\]
Now minus 1 raise to \[7\], means
\[ \Rightarrow {\left( { - 1} \right)^7}\]
Now \[2\] raise to 2, means
\[ \Rightarrow {\left( 2 \right)^2}\].
Then minus 5 raise to \[3 \times \] minus 1 raise to \[7 \times 2\] raise to 2 becomes
\[ \Rightarrow {\left( { - 5} \right)^3} \times {\left( { - 1} \right)^7} \times {\left( 2 \right)^2}\]
Now we need to simplify this algebraic expression. We know the negative power of an odd number is negative only. That is \[ {\left( { - P} \right)^Q} = negative{\text{ }}value\], where P is any number and Q is an odd number.
\[ \Rightarrow \left( { - 125} \right) \times \left( { - 1} \right) \times \left( 4 \right)\]
\[ \Rightarrow \left( { - 125} \right) \times \left( { - 4} \right)\]
\[ \Rightarrow 500\].
Hence the required answer is 500.
Note:
Algebra helps in converting a mathematical statement into an equation. We know if we have ‘more’ or ‘sum’ in the given sentence we use the addition operation \[( + )\]. Similarly If we have ‘less’ or ‘difference’ we use subtraction \[( - )\]. If we have ‘quotient’ we use division operation \[( \div )\]. To define more generalized terms; we use algebra. It is a very vast branch of mathematics and is used in all the branches of mathematics like polynomial, linear equations, graphs, etc. and in daily life too.
We know that the product of two negative numbers results in a positive number. The product of a positive (negative) number and a negative (positive) number results in a negative number.
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