Advertisements
Advertisements
प्रश्न
If `2^x xx3^yxx5^z=2160,` find x, y and z. Hence, compute the value of `3^x xx2^-yxx5^-z.`
Advertisements
उत्तर
Given `2^x xx3^yxx5^z=2160`
First, find out the prime factorisation of 2160.
It can be observed that 2160 can be written as `2^4xx3^3xx5^1`
Also,
`2^x xx36yxx5^z=2^4xx3^3xx5^1`
⇒ x = 4, y = 3, z = 1
Therefore, the value of `3^x xx2^-yxx5^-z` is `3^4xx2^-3xx5^-1=81xx1/8xx1/5=81/40`
APPEARS IN
संबंधित प्रश्न
Simplify the following:
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
Solve the following equation for x:
`7^(2x+3)=1`
Solve the following equation for x:
`4^(2x)=1/32`
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)xx5^(2y-3)=225`
If 102y = 25, then 10-y equals
If (16)2x+3 =(64)x+3, then 42x-2 =
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
Simplify:
`(3/5)^4 (8/5)^-12 (32/5)^6`
