Advertisements
Advertisements
प्रश्न
Solve the following equation for x:
`7^(2x+3)=1`
उत्तर
`7^(2x+3)=1`
`rArr7^(2x+3)=7^0`
⇒ 2x + 3 = 0
⇒ 2x = -3
`rArrx=-3/2`
APPEARS IN
संबंधित प्रश्न
Solve the following equation for x:
`2^(5x+3)=8^(x+3)`
Solve the following equation for x:
`4^(2x)=1/32`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt2/sqrt3)^5(6/7)^2`
Prove that:
`(2^(1/2)xx3^(1/3)xx4^(1/4))/(10^(-1/5)xx5^(3/5))div(3^(4/3)xx5^(-7/5))/(4^(-3/5)xx6)=10`
If a and b are different positive primes such that
`(a+b)^-1(a^-1+b^-1)=a^xb^y,` find x + y + 2.
The product of the square root of x with the cube root of x is
The simplest rationalising factor of \[2\sqrt{5}-\]\[\sqrt{3}\] is
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
If \[\frac{5 - \sqrt{3}}{2 + \sqrt{3}} = x + y\sqrt{3}\] , then
Find:-
`32^(2/5)`
