For a fixed base $y$ , if the exponent decreases by $1$ , the number becomes:

y^{th}

of the previous number

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{\frac{1}{y}}^{th}

of the previous number

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y^{2}

of the previous number

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\frac{1}{y^{2}}^{th}

of the previous number

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Solution

The correct option is B ${\frac{1}{y}}^{th}$ of the previous number
For $y^{4}$ , when the exponent decreases by $1$ , it becomes $y^{4 - 1} = y^{3}$
Now, $\frac{y^{3}}{y^{4}} = \frac{1}{y}$
Hence, for a fixed base $y$ , if the exponent decreases by $1$ , the number becomes ${\frac{1}{y}}^{th}$ of the previous number.

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Question 2

In the following question, out of the four options, only one is correct. Write the correct answer:

For a fixed base, if the exponent decreases by 1, the number becomes:
(a)One-thenth of the previous number
(b)Ten times of the previous number
(c)Hundredth of the previous number
(d)Hundred times fo the previous number

Q. What does the original value of

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become when the exponent decreases by 1?

If a certain number when decreased by $21$ becomes $81$ , what is the number?

___

Q. Find the exponent of the first term in

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For a fixed base y, if the exponent decreases by 1, the number becomes:

The correct option is B 1yth of the previous numberFor y4, when the exponent decreases by 1, it becomes y4−1=y3 Now, y3y4=1y Hence, for a fixed base y, if the exponent decreases by 1, the number becomes 1yth of the previous number.

For a fixed base $y$ , if the exponent decreases by $1$ , the number becomes: