Answer:
The voltage in exponential form would be [tex](\frac{3}{5})^3[/tex].
The original voltage in the circuit is [tex]\frac{3}{5}[/tex] of a volt.
Step-by-step explanation:
Let x represent the original voltage.
We have been given that the voltage in an electrical circuit is multiplied by itself each time it is reduced. The voltage is 27/125 of a volt and it has been reduced 3 times.
The original voltage multiplied by itself for 3 times would be [tex]x\cdot x\cdot x=x^3[/tex].
Now, we will equate [tex]x^3[/tex] with 27/125 as:
[tex]x^3=\frac{27}{125}[/tex]
We know that 27 is cube of 3 and 125 is cube of 5, so we can represent our equation as:
[tex]x^3=\frac{3^3}{5^3}[/tex]
[tex]x^3=(\frac{3}{5})^3[/tex]
Therefore, the voltage in exponential form would be [tex](\frac{3}{5})^3[/tex]
Now, we will take cube root of both sides to find original voltage in the circuit as:
[tex]\sqrt[3]{x^3} =\sqrt[3]{(\frac{3}{5})^3}[/tex]
Using property [tex]\sqrt[n]{a^n} =a[/tex], we will get:
[tex]x =\frac{3}{5}[/tex]
Therefore, the original voltage in the circuit is [tex]\frac{3}{5}[/tex] of a volt.
Answer: (3/5)^3 in exponential form and 3/5 a volt
Step-by-step explanation:
