How do you simplify ${( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2}$ ?
Answer
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Hint: When a number is raised to the power “n”, it means that the number is multiplied with itself “n” times, for example, let ${a^n}$ be an exponential function, it means that the number “a” is multiplied with itself “n” times. In this question, several exponential functions are combined to form one exponential function. $( - 3{d^2}f)$ is raised to the power 3, that is, $( - 3{d^2}f)$ is multiplied with itself 3 times, then the number obtained is raised to the power 2, that is, the number obtained is multiplied with itself 2 times. We know that the exponential functions obey certain rules called the laws of exponents. We will simplify this equation by using the knowledge of these laws.
Complete step-by-step solution:
We have to simplify ${( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2}$
We know that ${({a^x})^y} = {a^{x \times y}}$ .
So, ${[{( - 3{d^2}f)^3}]^2} = {( - 3{d^2}f)^{3 \times 2}} = {( - 3{d^2}f)^6}$
We know that ${(ab)^m} = {a^m}{b^m}$ , so we get –
$
\Rightarrow {( - 3{d^2}f)^6} = {( - 3)^6}{({d^2})^6}{f^6} \\
\Rightarrow ( - 3{d^2}f) = 729{d^{12}}{f^6} \\
$
And ${( - 3{d^2}{f^3}g)^2} = {( - 3)^2}{({d^2})^2}{({f^3})^2}{g^2} = 9{d^4}{f^6}{g^2}$
So, we get –
$
\Rightarrow {( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2} = 9{d^4}{f^6}{g^2} \times 729{d^{12}}{f^6} \\
\Rightarrow {( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2} = 9 \times 729 \times {d^4} \times {d^{12}} \times {f^6} \times {f^6} \times {g^2} \\
$
We also know that ${a^n} \times {a^m} = {a^{n + m}}$ , so we get –
$
\Rightarrow {( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2} = 6561{d^{4 + 12}}{f^{6 + 6}}{g^2} \\
\Rightarrow {( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2} = 6561{d^{16}}{f^{12}}{g^2} \\
$
Hence, the simplified form of ${( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2}$ is $6561{d^{16}}{f^{12}}{g^2}$ .
Note: In this question, we are given an equation that is in exponential form. A number is said to be in exponential form when the number is raised to some power, for example ${a^x}$ is an exponential function. There are several laws of exponents like the addition of two exponential functions, subtraction of two exponential functions, etc. The base of the exponential functions should be the same is a necessary condition for applying these laws. Thus, similar questions can be solved by using the laws of exponents.
Complete step-by-step solution:
We have to simplify ${( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2}$
We know that ${({a^x})^y} = {a^{x \times y}}$ .
So, ${[{( - 3{d^2}f)^3}]^2} = {( - 3{d^2}f)^{3 \times 2}} = {( - 3{d^2}f)^6}$
We know that ${(ab)^m} = {a^m}{b^m}$ , so we get –
$
\Rightarrow {( - 3{d^2}f)^6} = {( - 3)^6}{({d^2})^6}{f^6} \\
\Rightarrow ( - 3{d^2}f) = 729{d^{12}}{f^6} \\
$
And ${( - 3{d^2}{f^3}g)^2} = {( - 3)^2}{({d^2})^2}{({f^3})^2}{g^2} = 9{d^4}{f^6}{g^2}$
So, we get –
$
\Rightarrow {( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2} = 9{d^4}{f^6}{g^2} \times 729{d^{12}}{f^6} \\
\Rightarrow {( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2} = 9 \times 729 \times {d^4} \times {d^{12}} \times {f^6} \times {f^6} \times {g^2} \\
$
We also know that ${a^n} \times {a^m} = {a^{n + m}}$ , so we get –
$
\Rightarrow {( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2} = 6561{d^{4 + 12}}{f^{6 + 6}}{g^2} \\
\Rightarrow {( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2} = 6561{d^{16}}{f^{12}}{g^2} \\
$
Hence, the simplified form of ${( - 3{d^2}{f^3}g)^2}{[{( - 3{d^2}f)^3}]^2}$ is $6561{d^{16}}{f^{12}}{g^2}$ .
Note: In this question, we are given an equation that is in exponential form. A number is said to be in exponential form when the number is raised to some power, for example ${a^x}$ is an exponential function. There are several laws of exponents like the addition of two exponential functions, subtraction of two exponential functions, etc. The base of the exponential functions should be the same is a necessary condition for applying these laws. Thus, similar questions can be solved by using the laws of exponents.
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