
Find the value of ${{\left( 256 \right)}^{0.16}}\times {{\left( 256 \right)}^{0.09}}$
A. 4
B. 16
C. 64
D. 256.25
Answer
568.2k+ views
Hint: Here in this question we need to find the value of${{\left( 256 \right)}^{0.16}}\times {{\left( 256 \right)}^{0.09}}$ . We will initially simplify the expression using simple basic exponent formulae ${{a}^{x}}.{{a}^{y}}={{a}^{x+y}}$and ${{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}$. The factors of 256 are \[1,2,4,8,16,256,128,32,64\] because
$256=1\times 256=2\times 128=4\times 64=8\times 32=16\times 16$. We will use this for further simplifications which will lead us to the answer.
Complete step by step answer:
The value of${{\left( 256 \right)}^{0.16}}\times {{\left( 256 \right)}^{0.09}}$.
As we know that from the basic concept of exponents that the product of a raised to the power x and a raised to the power y is equal to a raised to the power of sum of x and y where x and y are positive integers and a is a real number.
Since we know that${{a}^{x}}.{{a}^{y}}={{a}^{x+y}}$.
This can be written as${{\left( 256 \right)}^{0.16+0.09}}$.
As we know that 0.16+0.09=0.25 we can simply write this as${{\left( 256 \right)}^{0.25}}$.
Since $0.25=\dfrac{1}{4}$ we can write this simply as${{\left( 256 \right)}^{\dfrac{1}{4}}}$.
As we know that from the basic concept of exponents that the value of a raised to the power $\dfrac{1}{n}$ is equal to the value of a raised to the power n where n positive integer and a is a real number.
As we know that ${{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}$this can be written as${{\left( 256 \right)}^{\dfrac{1}{4}}}=\sqrt[4]{256}$.
The factors of 256 are \[1,2,4,8,16,256,128,32,64\]because
$256=1\times 256=2\times 128=4\times 64=8\times 32=16\times 16$.
We will use the grouping $256=4\times 4\times 4\times 4$and get the desired values.
Since we know that $256={{4}^{4}}$we can write this as $\sqrt[4]{256}=\sqrt[4]{{{4}^{4}}}$
Hence we can say that$\sqrt[4]{{{4}^{4}}}=4$.
So we will end up having the conclusion that the value of ${{\left( 256 \right)}^{0.16}}\times {{\left( 256 \right)}^{0.09}}$ is 4.
So, the correct answer is “Option A”.
Note: While solving these types of questions we should remember the properties of exponents. Let us discuss the different basic properties of exponents. There are 8 basic properties when a and b are real numbers and x and y and n are positive integers.
1. ${{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}$
2. ${{a}^{x}}.{{a}^{y}}={{a}^{x+y}}$
3. ${{a}^{0}}=1,\forall a\ne 0$
4. ${{a}^{x}}\div {{a}^{y}}={{a}^{x-y}}$
5. ${{\left( {{a}^{x}} \right)}^{y}}={{a}^{x\times y}}$
6. ${{\left( \dfrac{a}{b} \right)}^{n}}={{a}^{n}}\times {{b}^{-n}}$
7. ${{\left( a\times b \right)}^{n}}={{a}^{n}}\times {{b}^{n}}$
8. ${{\left( \dfrac{a}{b} \right)}^{-n}}={{\left( \dfrac{b}{a} \right)}^{n}}$
$256=1\times 256=2\times 128=4\times 64=8\times 32=16\times 16$. We will use this for further simplifications which will lead us to the answer.
Complete step by step answer:
The value of${{\left( 256 \right)}^{0.16}}\times {{\left( 256 \right)}^{0.09}}$.
As we know that from the basic concept of exponents that the product of a raised to the power x and a raised to the power y is equal to a raised to the power of sum of x and y where x and y are positive integers and a is a real number.
Since we know that${{a}^{x}}.{{a}^{y}}={{a}^{x+y}}$.
This can be written as${{\left( 256 \right)}^{0.16+0.09}}$.
As we know that 0.16+0.09=0.25 we can simply write this as${{\left( 256 \right)}^{0.25}}$.
Since $0.25=\dfrac{1}{4}$ we can write this simply as${{\left( 256 \right)}^{\dfrac{1}{4}}}$.
As we know that from the basic concept of exponents that the value of a raised to the power $\dfrac{1}{n}$ is equal to the value of a raised to the power n where n positive integer and a is a real number.
As we know that ${{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}$this can be written as${{\left( 256 \right)}^{\dfrac{1}{4}}}=\sqrt[4]{256}$.
The factors of 256 are \[1,2,4,8,16,256,128,32,64\]because
$256=1\times 256=2\times 128=4\times 64=8\times 32=16\times 16$.
We will use the grouping $256=4\times 4\times 4\times 4$and get the desired values.
Since we know that $256={{4}^{4}}$we can write this as $\sqrt[4]{256}=\sqrt[4]{{{4}^{4}}}$
Hence we can say that$\sqrt[4]{{{4}^{4}}}=4$.
So we will end up having the conclusion that the value of ${{\left( 256 \right)}^{0.16}}\times {{\left( 256 \right)}^{0.09}}$ is 4.
So, the correct answer is “Option A”.
Note: While solving these types of questions we should remember the properties of exponents. Let us discuss the different basic properties of exponents. There are 8 basic properties when a and b are real numbers and x and y and n are positive integers.
1. ${{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}$
2. ${{a}^{x}}.{{a}^{y}}={{a}^{x+y}}$
3. ${{a}^{0}}=1,\forall a\ne 0$
4. ${{a}^{x}}\div {{a}^{y}}={{a}^{x-y}}$
5. ${{\left( {{a}^{x}} \right)}^{y}}={{a}^{x\times y}}$
6. ${{\left( \dfrac{a}{b} \right)}^{n}}={{a}^{n}}\times {{b}^{-n}}$
7. ${{\left( a\times b \right)}^{n}}={{a}^{n}}\times {{b}^{n}}$
8. ${{\left( \dfrac{a}{b} \right)}^{-n}}={{\left( \dfrac{b}{a} \right)}^{n}}$
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