0.01024 times (0.4)^{9} = (0.4)^{?} times (0. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equation: $0.01024 	imes (0.4)^{9} = (0.4)^{x} 	imes (0.0256)^{3}$Express the decimals as powers of $4$ and $10$:$0.01024 = \frac{1024}{10

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given equation: $0.01024 	imes (0.4)^{9} = (0.4)^{x} 	imes (0.0256)^{3}$Express the decimals as powers of $4$ and $10$:$0.01024 = \frac{1024}{100000} = \frac{4^{5}}{10^{5}}$$0.0256 = \frac{256}{10000} = \frac{4^{4}}{10^{4}}$Substitute these into the equation:$\frac{4^{5}}{10^{5}} 	imes (0.4)^{9} = (0.4)^{x} 	imes \left(\frac{4^{4}}{10^{4}}ight)^{3}$Since $0.4 = \frac{4}{10}$,we have:$\frac{4^{5}}{10^{5}} 	imes \left(\frac{4}{10}ight)^{9} = \left(\frac{4}{10}ight)^{x} 	imes \left(\left(\frac{4}{10}ight)^{4}ight)^{3}$$\left(\frac{4}{10}ight)^{5} 	imes \left(\frac{4}{10}ight)^{9} = \left(\frac{4}{10}ight)^{x} 	imes \left(\frac{4}{10}ight)^{12}$Using the law of exponents $a^{m} 	imes a^{n} = a^{m+n}$:$\left(\frac{4}{10}ight)^{5+9} = \left(\frac{4}{10}ight)^{x+12}$$\left(\frac{4}{10}ight)^{14} = \left(\frac{4}{10}ight)^{x+12}$Equating the exponents:$14 = x + 12$$x = 14 - 12 = 2$

$0.01024 \times (0.4)^{9} = (0.4)^{?} \times (0.0256)^{3}$

Similar Questions

How many natural numbers are there between $23$ and $100$ which are exactly divisible by $6$?

The greatest number among $3^{50}, 4^{40}, 5^{30}$ and $6^{20}$ is

Find the nearest number to $12199$ which is exactly divisible by the product of the first four prime numbers.

$A$ and $B$ are positive integers. If $A + B + AB = 65$,then what is the difference between $A$ and $B$ (given $A, B \leq 15$)?

$n$ is a whole number which when divided by $4$ gives the remainder $3$. The remainder when $2n$ is divided by $4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module