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Question

Question: How do you rewrite the exponential expression as a radical expression \({b^{ - \dfrac{4}{7}}}\)?...

How do you rewrite the exponential expression as a radical expression b47{b^{ - \dfrac{4}{7}}}?

Explanation

Solution

In this question we have an exponential expression which has a negative power which is in terms of a fraction therefore, we will use two exponential properties to simplify the expression and write it in the radical form.

Formula used:
1ab=ab\dfrac{1}{{{a^b}}} = {a^{ - b}}
abc=(ab)1c=pbc{a^{\dfrac{b}{c}}} = {({a^b})^{\dfrac{1}{c}}} = \sqrt[c]{{{p^b}}}

Complete step-by-step answer:
We have the expression as:
b47\Rightarrow {b^{ - \dfrac{4}{7}}}
Now since the exponent is in the negative we know, the inverse of the number, also called the reciprocal of the number is the number dividing 11 , for example the reciprocal of is 1a\dfrac{1}{a} , and it can also be expressed in terms of power as a1{a^{ - 1}}.
Therefore, on using this property, we get:
1b47\Rightarrow \dfrac{1}{{{b^{\dfrac{4}{7}}}}}
Now the denominator is in term of exponent of a fraction, on using the formula abc=(ab)1c=pbc{a^{\dfrac{b}{c}}} = {({a^b})^{\dfrac{1}{c}}} = \sqrt[c]{{{p^b}}}, we get:
1(p4)17\Rightarrow \dfrac{1}{{{{({p^4})}^{\dfrac{1}{7}}}}}
Which could be written as:

1b47 \Rightarrow \dfrac{1}{{\sqrt[7]{{{b^4}}}}}, which is the required solution.

Note:
It is to be noted that the function can be read out as ‘11 divided by the 7th7th root of bb raised to 44’. This means that whatever is the solution when the term bb is raised to the power of 44, we have to take the seventh root of the number, which means the number which has to be multiplied 77 times to get b4{b^4}.
Exponents are used to write the similar terms in multiplication in a simple format. Some exponential numbers are too big to be calculated directly, therefore to solve them logarithm is used.
Logarithm is used in exponents to convert the exponential term in the form of multiplication. The antilog of the term is then taken which is the inverse of taking log, to get the required solution back after simplification.