Advertisements
Advertisements
Question
Solve the following equation:
`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`
Advertisements
Solution
`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`
`rArr4^(2x)=(sqrt8)^2` and `(root3 16)^(-6/y)=(sqrt8)^2`
`rArr4^(2x)=(8^1/2xx2)` and `(16^(1/3xx-6/y))=(8^1/2xx2)`
`rArr4^(2x)=8` and `(16^(-2/y))=8`
`rArr2^(4x)=2^3` and `(2^(-8/y))=2^3`
`rArr4x=3` and `-8/y=3`
`rArrx=3/4` and `y=-8/3`
APPEARS IN
RELATED QUESTIONS
If a = 3 and b = -2, find the values of :
(a + b)ab
Simplify the following:
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
If `1176=2^a3^b7^c,` find a, b and c.
Prove that:
`((0.6)^0-(0.1)^-1)/((3/8)^-1(3/2)^3+((-1)/3)^-1)=(-3)/2`
If `27^x=9/3^x,` find x.
Find the value of x in the following:
`(13)^(sqrtx)=4^4-3^4-6`
State the product law of exponents.
If 9x+2 = 240 + 9x, then x =
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
