
How do you solve ${2^x} = \dfrac{1}{{32}}?$
Answer
554.7k+ views
Hint:First simplify the given number into its factors and then use law of indices for multiplication and then use law of indices for negative powers and at last compare the exponents of the same bases in order to find the solution for the given equation.
Formula used:
Law of indices for multiplication and for negative powers is given as follows:
${a^m} \times {a^n} = {a^{m + n}}$
$\dfrac{1}{{{a^m}}} = {a^{ - m}}$
Complete step by step answer:
In order to find the solution for the given equation ${2^x} = \dfrac{1}{{32}}$, first we will simplify the constants present in the equation into their factors.
We can see that in the given equation ${2^x} = \dfrac{1}{{32}}$, only $\dfrac{1}{{32}}$ is the constant, so finding its factors,
We can write $32\;as\;2 \times 2 \times 2 \times 2 \times 2$
That is $\dfrac{1}{{32}} = \dfrac{1}{{2 \times 2 \times 2 \times 2 \times 2}}$
Now using the law of indices for multiplication which says if two or more numbers of equal base are multiplying together then their product can be written as the base to the power of sum of all the powers, mathematically it is ${a^m} \times {a^n} = {a^{m + n}}$
$\dfrac{1}{{2 \times 2 \times 2 \times 2 \times 2}} = \dfrac{1}{{{2^{1 + 1 + 1 + 1 + 1}}}} = \dfrac{1}{{{2^5}}}$
Again using the law of indices but this time for negative powers which can be understood mathematically as
$\dfrac{1}{{{a^m}}} = {a^{ - m}}$
Applying this to simplify the constant further, we will get
$\dfrac{1}{{{2^5}}} = {2^{ - 5}}$
So we have simplified the constant, now putting this in the equation,
$
\Rightarrow {2^x} = \dfrac{1}{{32}} \\
\Rightarrow {2^x} = {2^{ - 5}} \\ $
Since bases in both the sides (i.e. L.H.S. and R.H.S.) are equal so their exponents must have to be equal.
$\therefore x = - 5$
So $x = - 5$ is the solution for the equation ${2^x} = \dfrac{1}{{32}}$.
Note: In order to solve this type of equation always try to write the constants in the equation in factors of the term carrying variable in the equation, so that you can compare their exponents after simplifying the constants.
Formula used:
Law of indices for multiplication and for negative powers is given as follows:
${a^m} \times {a^n} = {a^{m + n}}$
$\dfrac{1}{{{a^m}}} = {a^{ - m}}$
Complete step by step answer:
In order to find the solution for the given equation ${2^x} = \dfrac{1}{{32}}$, first we will simplify the constants present in the equation into their factors.
We can see that in the given equation ${2^x} = \dfrac{1}{{32}}$, only $\dfrac{1}{{32}}$ is the constant, so finding its factors,
We can write $32\;as\;2 \times 2 \times 2 \times 2 \times 2$
That is $\dfrac{1}{{32}} = \dfrac{1}{{2 \times 2 \times 2 \times 2 \times 2}}$
Now using the law of indices for multiplication which says if two or more numbers of equal base are multiplying together then their product can be written as the base to the power of sum of all the powers, mathematically it is ${a^m} \times {a^n} = {a^{m + n}}$
$\dfrac{1}{{2 \times 2 \times 2 \times 2 \times 2}} = \dfrac{1}{{{2^{1 + 1 + 1 + 1 + 1}}}} = \dfrac{1}{{{2^5}}}$
Again using the law of indices but this time for negative powers which can be understood mathematically as
$\dfrac{1}{{{a^m}}} = {a^{ - m}}$
Applying this to simplify the constant further, we will get
$\dfrac{1}{{{2^5}}} = {2^{ - 5}}$
So we have simplified the constant, now putting this in the equation,
$
\Rightarrow {2^x} = \dfrac{1}{{32}} \\
\Rightarrow {2^x} = {2^{ - 5}} \\ $
Since bases in both the sides (i.e. L.H.S. and R.H.S.) are equal so their exponents must have to be equal.
$\therefore x = - 5$
So $x = - 5$ is the solution for the equation ${2^x} = \dfrac{1}{{32}}$.
Note: In order to solve this type of equation always try to write the constants in the equation in factors of the term carrying variable in the equation, so that you can compare their exponents after simplifying the constants.
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