Advertisements
Advertisements
प्रश्न
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
Advertisements
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
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Simplify:
`(16^(-1/5))^(5/2)`
Prove that:
`(2^(1/2)xx3^(1/3)xx4^(1/4))/(10^(-1/5)xx5^(3/5))div(3^(4/3)xx5^(-7/5))/(4^(-3/5)xx6)=10`
Show that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
The value of \[\left\{ 2 - 3 (2 - 3 )^3 \right\}^3\] is
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
The simplest rationalising factor of \[\sqrt[3]{500}\] is
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
