Advertisements
Advertisements
प्रश्न
If 102y = 25, then 10-y equals
विकल्प
- \[- \frac{1}{5}\]
- \[\frac{1}{50}\]
- \[\frac{1}{625}\]
- \[\frac{1}{5}\]
Advertisements
उत्तर
We have to find the value of `10^-y`
Given that, `10^(2y) = 25` therefore,
`10^(2y) = 25`
`(10^y)^2 = 5^2`
`(10^y)^(2 xx 1/2)= 5^(2 xx 1/2)`
`(10^y)^(2 xx 1/2)= 5^(2 xx 1/2)`
`10^y/1 = 5/1`
`1/5 =1/10^y`
`1/5 =10^-y`
APPEARS IN
संबंधित प्रश्न
Simplify the following
`(4ab^2(-5ab^3))/(10a^2b^2)`
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Simplify:
`(0.001)^(1/3)`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
If \[\frac{3^{5x} \times {81}^2 \times 6561}{3^{2x}} = 3^7\] then x =
If \[\sqrt{2^n} = 1024,\] then \[{3^2}^\left( \frac{n}{4} - 4 \right) =\]
If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=
Find:-
`125^(1/3)`
Find:-
`125^((-1)/3)`
