Advertisements
Advertisements
Question
Simplify the following:
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
Advertisements
Solution
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
`=(5xx(5^2)^(n+1)-(5^2)xx5^(2n))/(5xx5^(2n+3)-(5^2)^(n+1))`
`=(5xx(5^(2n+2))-(5^2)xx5^(2n))/(5xx5^(2n+3)-5^(2n+2))`
`=(5^(1+2n+2)-5^(2+2n))/(5^(1+2n+3)-5^(2n+2))`
`=(5^(2+2n)(5-1))/(5^(2+2n)(5^2-1))`
`=(5-1)/((5^2)-1)`
`=(5-1)/(25-1)`
`=4/24`
`=1/6`
APPEARS IN
RELATED QUESTIONS
Solve the following equation for x:
`4^(2x)=1/32`
Solve the following equations for x:
`2^(2x)-2^(x+3)+2^4=0`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
If a and b are distinct primes such that `root3 (a^6b^-4)=a^xb^(2y),` find x and y.
If a and b are different positive primes such that
`(a+b)^-1(a^-1+b^-1)=a^xb^y,` find x + y + 2.
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 =
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
