Steps: 1 Mark
Answer: 1 Mark
$4^5=\left(4\right)\times \left(4\right)\times \left(4\right)\times \left(4\right)\times \left(4\right)$
$\Rightarrow 4^5=1024$
$5^4=\left(5\right)\times \left(5\right)\times \left(5\right)\times \left(5\right)$
$\Rightarrow 5^4=625$
The difference between $4^5$ and $5^4$ is
$4^5-5^4=1024-625$
$\Rightarrow 4^5-5^4=399$
The difference between $4^5$ and $5^4$ is $399$.

Application: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Consider,
$\frac{8^4\times {14}^3\times 5\times 3^4}{4^5\times 7^3\times 6^2}$
$8^4=(2^3{)}^4=2^{12}$
${14}^3=(2\times 7{)}^3=2^3\times 7^3$
$4^5=(2^2{)}^5=2^{10}$
$6^2=(2\times 3{)}^2=2^2\times 3^2$
On substituting the values we get:
$\frac{2^{12}\times 2^3\times 7^3\times 5\times 3^4}{2^{10}\times 7^3\times 2^2\times 3^2}$
$=2^{12+3-10-2}\times 7^{3-3}\times 5\times 3^{4-2}$ [$\because \frac{a^m}{a^n}=a^{m-n}$]
$=2^3\times 5\times 3^2$   [$\because a^0=1$]
$=8\times 5\times 9$
$=360$

Concept: 1 Mark
Application: 1 Mark
Answer: 1 Mark
Let the number be $x$.
Given that,
$\left(x\right)\times \left(25\right)=78125$
$x=3125$
$\Rightarrow$ The required number is 3125
Now,
$3125=5\times 5\times 5\times 5\times 5$
$\Rightarrow 3125=5^5$
Now, if the number was multiplied by $(-5)\times \left(5\right)$ then the product$=(-5)\times \left(5\right)\times 3125$ = $-78125$

Conversion: 1 Mark
Difference: 1 Mark
Standard form: 1 Mark
As per the question
The amount shopkeeper A has = Rs 34265.
The amount shop keeper B has = 432500 paise $\Rightarrow$  Rs 4325.
Difference = Rs 34265 - Rs 4325 = Rs 29940.
The standard form of Rs 29940.
$29940=2.994\times {10}^4$
The difference in rupees is $2.994\times {10}^4$

Formula: 1 Mark
Application: 1 Mark
Answer: 2 Marks
Given that
The ball is released from a height of $xm$
Height reached after first bounce $=\frac{x}{2}$
Height reached after second bounce $=\frac{x}{4}=\frac{x}{2^2}$
Height reached after third bounce $=\frac{x}{8}=\frac{x}{2^3}$
Height reached after fourth bounce $=\frac{x}{16}=\frac{x}{2^4}$
Height reached after fifth bounce $=\frac{x}{32}=\frac{x}{2^5}$
Height reached after sixth bounce $=\frac{x}{64}=\frac{x}{2^6}$
Height reached after seventh bounce $=\frac{x}{128}=\frac{x}{2^7}$
Height reached after eight bounce $=\frac{x}{256}=\frac{x}{2^8}$
$\therefore$ Product of first 9 heights $=x\times \frac{x}{2}\times \frac{x}{2^2}\times \frac{x}{2^3}\times \frac{x}{2^4}\times \frac{x}{2^5}\times \frac{x}{2^6}\times \frac{x}{2^7}\times \frac{x}{2^8}$
$=\frac{x^{(1+1+1+1+1+1+1+1+1)}}{2^{(1+2+3+4+5+6+7+8)}}$
$(\because a^b\times a^c=a^{b+c})$
$=\frac{x^9}{2^{36}}$
The product of heights after 8 bounces $=\frac{x^9}{2^{36}}$

Step: 2 Marks
Application: 1 Mark
Answer: 1 Mark
The area of land owned by X is
$=3^4{m}^2=81m^2$
The value of given property is
$=\left(81\right)\times \left(50\right)=Rs.4050$
The area of land owned by Y is 
$4^3{m}^2=64m^2$
The value of given property is
$=\left(64\right)\times \left(50\right)=Rs.3200$
$\Rightarrow$ Value of property owned by X is greater than property owned by Y.