Advertisements
Advertisements
Question
If `2^x xx3^yxx5^z=2160,` find x, y and z. Hence, compute the value of `3^x xx2^-yxx5^-z.`
Advertisements
Solution
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
RELATED QUESTIONS
Solve the following equation for x:
`7^(2x+3)=1`
If `5^(3x)=125` and `10^y=0.001,` find x and y.
If a and b are distinct primes such that `root3 (a^6b^-4)=a^xb^(2y),` find x and y.
Show that:
`((a+1/b)^mxx(a-1/b)^n)/((b+1/a)^mxx(b-1/a)^n)=(a/b)^(m+n)`
`(2/3)^x (3/2)^(2x)=81/16 `then x =
The value of \[\left\{ 8^{- 4/3} \div 2^{- 2} \right\}^{1/2}\] is
If (16)2x+3 =(64)x+3, then 42x-2 =
If x= \[\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\] and y = \[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\] , then x2 + y +y2 =
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
Find:-
`125^(1/3)`
