Advertisements
Advertisements
Question
If `5^(3x)=125` and `10^y=0.001,` find x and y.
Advertisements
Solution
It is given that `5^(3x)=125` and `10^y=0.001`.
Now,
`5^(3x)=125`
`rArr5^(3x)=5^3`
`rArr3x = 3`
x = 1
And,
`10^y=0.001`
`rArr10^y=1/1000`
`rArr10^y=10^-3`
⇒ y = -3
hence, the value of x and yare 1 and -3, respectively.
APPEARS IN
RELATED QUESTIONS
Solve the following equation for x:
`2^(5x+3)=8^(x+3)`
Simplify:
`(sqrt2/5)^8div(sqrt2/5)^13`
Prove that:
`((0.6)^0-(0.1)^-1)/((3/8)^-1(3/2)^3+((-1)/3)^-1)=(-3)/2`
Show that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
If `x=2^(1/3)+2^(2/3),` Show that x3 - 6x = 6
Write the value of \[\sqrt[3]{125 \times 27}\].
If x = 2 and y = 4, then \[\left( \frac{x}{y} \right)^{x - y} + \left( \frac{y}{x} \right)^{y - x} =\]
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
Simplify:
`(9^(1/3) xx 27^(-1/2))/(3^(1/6) xx 3^(- 2/3))`
