Advertisements
Advertisements
प्रश्न
Solve the following equation for x:
`4^(2x)=1/32`
उत्तर
`4^(2x)=1/32`
`rArr(2^2)^(2x)=1/2^5`
`rArr2^(4x)xx2^5=1`
`rArr2^(4x+5)=2^0`
⇒ 4x + 5 = 0
⇒ 4x = -5
`rArr x=-5/4`
APPEARS IN
संबंधित प्रश्न
Prove that:
`(a+b+c)/(a^-1b^-1+b^-1c^-1+c^-1a^-1)=abc`
If `1176=2^a3^b7^c,` find a, b and c.
Given `4725=3^a5^b7^c,` find
(i) the integral values of a, b and c
(ii) the value of `2^-a3^b7^c`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
Write \[\left( 625 \right)^{- 1/4}\] in decimal form.
`(2/3)^x (3/2)^(2x)=81/16 `then x =
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=
