Advertisements
Advertisements
प्रश्न
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
उत्तर
We have to find the value of `x/y` for `3^(x-1) = 9.4^(y+2) = 64`
So,
`3^(x-4) = 3 ^2`
By equating the exponent we get
x-1=2
x=2+1
x=3
Let’s take `4^(y+2) = 64`
`4^(y+2) = 4^3`
By equating the exponent we get
y+2 = 3
y=3-2
y=1
By substituting x=3,y=1 in `x/y` we get `3/1`
Hence the value of `x/y` is 3.
APPEARS IN
संबंधित प्रश्न
Simplify the following:
`(5^(n+3)-6xx5^(n+1))/(9xx5^x-2^2xx5^n)`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^((-2)/3)y^((-1)/2))^2`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Simplify:
`root(lm)(x^l/x^m)xxroot(mn)(x^m/x^n)xxroot(nl)(x^n/x^l)`
State the power law of exponents.
Write \[\left( \frac{1}{9} \right)^{- 1/2} \times (64 )^{- 1/3}\] as a rational number.
The value of 64-1/3 (641/3-642/3), is
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
The positive square root of \[7 + \sqrt{48}\] is
