Advertisements
Advertisements
प्रश्न
If `5^(3x)=125` and `10^y=0.001,` find x and y.
Advertisements
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
Simplify the following
`(2x^-2y^3)^3`
Prove that:
`9^(3/2)-3xx5^0-(1/81)^(-1/2)=15`
Prove that:
`((0.6)^0-(0.1)^-1)/((3/8)^-1(3/2)^3+((-1)/3)^-1)=(-3)/2`
Solve the following equation:
`3^(x+1)=27xx3^4`
If `a=x^(m+n)y^l, b=x^(n+l)y^m` and `c=x^(l+m)y^n,` Prove that `a^(m-n)b^(n-l)c^(l-m)=1`
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
For any positive real number x, find the value of \[\left( \frac{x^a}{x^b} \right)^{a + b} \times \left( \frac{x^b}{x^c} \right)^{b + c} \times \left( \frac{x^c}{x^a} \right)^{c + a}\].
The value of \[\left\{ \left( 23 + 2^2 \right)^{2/3} + (140 - 19 )^{1/2} \right\}^2 ,\] is
The simplest rationalising factor of \[\sqrt[3]{500}\] is
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
