${10}^{-5}$
= $\frac{1}{{10}^5}$
= $\frac{1}{100000}$
When a number is multiplied with its multiplicative inverse, the result is 1. To make the given number, which is $\frac{1}{100000}$, equal to 1, we need to multiply it with 100000, which is the reciprocal of $\frac{1}{100000}$.
Therefore, 100000, or ${10}^5$ is the multiplicative inverse of ${10}^{-5}$.

$4\times {10}^3+3\times 1+1\times {10}^{-1}+6\times {10}^0=(4\times 1000)+(3\times 1)+\frac{1}{10}+(6\times 1)=4000+3+0.1+6=4009.1$

$4.95\times {10}^5=4.95\times 100000=495000$

Note that $a^0=1$ for any $a\ne 0$
$\text{Thus, the value of}{1000}^{\circ }\text{is 1.}$

$(-2{)}^2\times (-5{)}^3=50m$
$\Rightarrow 4\times (-125)=50m$
$\Rightarrow -500=50m$
$\Rightarrow m=-10$

 $4^x=64$
 $\Rightarrow 4^x$=$4^3$            (as 64 is $4^3$ )
 $\Rightarrow x=3(\because a^m=a^{n}ifandonlyifm=nfora\ne 0,1,-1)$
 Since $x=3$,  $2^{2x}$ = 64 ,  and 2x =6

The usual forms of the expressions $9\times {10}^{-2}$ and $9\times {10}^3$ are given by 0.09 and 9000.
Thus, 9000 + 0.09 = 9000.09.

$x^5$ × $x^4$, for x = 2
Substituting 2 in place of $x$
= $2^5$× $2^4$
= $2^9(\because a^m\times a^n=a^{m+n})$ = 512

We know that $a^{-m}=\frac{1}{a^m}$ for any non-zero integer $a$ and integer $m.$
$⟹7^{-3}=\frac{1}{7^3}$
                 $=\frac{1}{7\times 7\times 7}=\frac{1}{343}$

Multiplicative inverse of a non-zero $x$ is $\frac{1}{x}$; so that when these two are multiplied, the product value is 1. Therefore, the multiplicative inverse of ${10}^{-7}$should be a number which when multiplied with ${10}^{-7}$ gives 1.
Let the multiplicative inverse of ${10}^{-7}$ be $x.$
Then, ${10}^{-7}\times x=1.$
$\Rightarrow x=\frac{1}{{10}^{-7}}={10}^7$
$(\because a^{-n}=\frac{1}{a^n}⟹\frac{1}{a^{-n}}=\frac{1}{\frac{1}{a^n}}=a^n)$
Thus, the multiplicative inverse of ${10}^{-7}$ is ${10}^7.$