Question

answer this question please​

Answer 1

Answer:

-1

Step-by-step explanation:

First we will use laws of indices; the third one

so it would be 2 raised to -1 into -1 minus 3 raised to -1 into -1

(since, - into - is equal to plus)

2 raised to 1 minus 3 raised to 1

2 - 3 = -1

:. 2 - 3 = -1

Answer 2

Working out:

In this question, we need to find the result of the above exponent. We can do this by using different exponent and index laws which I will list out between the steps.

GiveN:

• $\sf{( {2}^{ - 1} - {3}^{ - 1} ) {}^{ - 1}}$

Solution:

Let's start solving the exponents of 2 and 3 by using the exponent law of $\sf{{a}^{-n} = \frac{1}{{a}^{n}}}$ .

$\sf{ \longrightarrow{ \bigg( \dfrac{1}{2} - \dfrac{1}{3} \bigg) {}^{ - 1} }}$

Taking LCM within the parentheses,

$\sf{ \longrightarrow{ \bigg( \dfrac{3 - 2}{6} \bigg) {}^{ - 1} }}$

$\sf{ \longrightarrow{ \bigg( \dfrac{1}{6} \bigg) {}^{ - 1} }}$

Now again using the same law of $\sf{{a}^{-n} = \frac{1}{{a}^{n}}}$ for getting the result,

$\sf{ \longrightarrow{6 {}^{1} }}$

And hence the result of the above exponent:

$\LARGE{\therefore{ \boxed{ \red{ \sf{6}}}}}$

And we are done !!