Advertisements
Advertisements
Question
Solve the following equation:
`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`
Advertisements
Solution
`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`
`rArr(2^3)^(x+1)=(2^4)^(y+2)` and `(1/2)^(3+x)=(1/2^2)^(3y)`
`rArr(2)^(3x+3)=(2)^(4y+8)` and `(1/2)^(3+x)=(1/2)^(6y)`
⇒ 3x + 3 = 4y + 8 and 3 + x = 6y
⇒ 3x - 4y = 8 - 3
⇒ 3x - 4y = 5 ...........(i)
Now,
3 + x = 6y
x = 6y - 3 ..............(ii)
Putting x = 6y - 3 in equation (i), we get
3(6y - 3) - 4y = 5
⇒ 18y - 9 - 4y = 5
⇒ 14y = 14
⇒ y = 1
Putting y = 1 in equation (ii) we get,
x = 6(1) - 3 = 3
APPEARS IN
RELATED QUESTIONS
Prove that:
`(x^a/x^b)^(a^2+ab+b^2)xx(x^b/x^c)^(b^2+bc+c^2)xx(x^c/x^a)^(c^2+ca+a^2)=1`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
Solve the following equation:
`sqrt(a/b)=(b/a)^(1-2x),` where a and b are distinct primes.
If a and b are different positive primes such that
`((a^-1b^2)/(a^2b^-4))^7div((a^3b^-5)/(a^-2b^3))=a^xb^y,` 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.
If (x − 1)3 = 8, What is the value of (x + 1)2 ?
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
Which one of the following is not equal to \[\left( \sqrt[3]{8} \right)^{- 1/2} ?\]
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
