Advertisements
Advertisements
प्रश्न
If `5^(3x)=125` and `10^y=0.001,` find x and y.
उत्तर
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
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`root5(243x^10y^5z^10)`
Simplify:
`(0.001)^(1/3)`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
Solve the following equation:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
If `2^x xx3^yxx5^z=2160,` find x, y and z. Hence, compute the value of `3^x xx2^-yxx5^-z.`
State the power law of exponents.
For any positive real number x, write the value of \[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\]
If x-2 = 64, then x1/3+x0 =
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
