Advertisements
Advertisements
प्रश्न
Solve the following equation for x:
`4^(2x)=1/32`
Advertisements
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`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Given `4725=3^a5^b7^c,` find
(i) the integral values of a, b and c
(ii) the value of `2^-a3^b7^c`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
If a, b, c are positive real numbers, then \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
Find:-
`32^(2/5)`
Simplify:
`11^(1/2)/11^(1/4)`
If `a = 2 + sqrt(3)`, then find the value of `a - 1/a`.
