Question

please tell with explanation​

Answer 1

Question:-

If $\sf{\sqrt{2^x}}$ = 32, then $\sf{\dfrac{x - 1}{x}}$ is equal to

(a) 9

(b) 0.9

(c) 10

(d) $\sf{\dfrac{10}{9}}$

Given:-

• $\sf{\sqrt{2^x} = 32}$

To Find:-

• The value of $\sf{\dfrac{x-1}{x}}$

Solution:-

Its given that

$\sf{\sqrt{2^x} = 32}$

Squaring both the sides,

$\sf{(\sqrt{2^x})^2 = [(2)^5]^2}$

According to law of exponents,

$\sf{[a^n]^m = a^{nm}}$

$\sf{2^x = 2^{10}}$

According to law of exponents,

$\sf{a^m = a^n}$

$\sf{m = n}$

It means that if the bases on the either sides of equal to sides are same then the power of both the bases will be equal.

Here,

$\sf{2^x = 2^{10}}$

Here,

The bases are same,

Hence,

$\sf{x = 10}$

Now,

We need to find the value of $\sf{\dfrac{x-1}{x}}$

Putting the value of x

$\sf{\dfrac{10-1}{10}}$

= $\sf{\dfrac{9}{10}}$

= 0.9

Hence the correct answer is option (b) 0.9.

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Explore more!!!

Some other important laws of exponents:-

• $\sf{a^m \times a^n = a^{m + n}}$
• $\sf{\dfrac{a^n}{a^m} = a^{n-m}}$

• $\sf{a^0 = 1}$
• $\sf{a^{-1} = \dfrac{1}{a}}$

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