Advertisements
Advertisements
प्रश्न
The value of m for which \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\] is
विकल्प
\[- \frac{1}{3}\]
\[\frac{1}{4}\]
-3
2
Advertisements
उत्तर
We have to find the value of m for \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\]
⇒ `[{1/(7^(2x-2))}^-1/3]^(1/4) = 7^m`
⇒ `[{1/(7^-4)}^(-1/3)]^(1/4) = 7^m`
⇒ `[{1/(7^(-4x(-1)/3)) }}^(1/4)= 7^m`
⇒ `[{1/(7^(4/3))}]^)1/4 = 7^m`
⇒ `[{1/(7^(4/3 xx1/4))}] = 7^m`
⇒ `[{1/(7^(4/3 xx1/4))}] = 7^m`
⇒ `[1/(7^(1/3))] = 7^m`
By using rational exponents `1/a^n = a^-n`
\[7^\frac{- 1}{3} = 7^m\]
Equating power of exponents we get `- 1/3 = m`
APPEARS IN
संबंधित प्रश्न
Simplify:
`(0.001)^(1/3)`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
Solve the following equation:
`3^(x+1)=27xx3^4`
Solve the following equation:
`sqrt(a/b)=(b/a)^(1-2x),` where a and b are distinct primes.
Simplify:
`(x^(a+b)/x^c)^(a-b)(x^(b+c)/x^a)^(b-c)(x^(c+a)/x^b)^(c-a)`
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
The simplest rationalising factor of \[\sqrt[3]{500}\] is
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
If `a = 2 + sqrt(3)`, then find the value of `a - 1/a`.
