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`
APPEARS IN
संबंधित प्रश्न
Simplify the following
`((4xx10^7)(6xx10^-5))/(8xx10^4)`
Simplify the following:
`(6(8)^(n+1)+16(2)^(3n-2))/(10(2)^(3n+1)-7(8)^n)`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
Solve the following equation:
`sqrt(a/b)=(b/a)^(1-2x),` where a and b are distinct primes.
The value of x − yx-y when x = 2 and y = −2 is
If 10x = 64, what is the value of \[{10}^\frac{x}{2} + 1 ?\]
Simplify:
`(1^3 + 2^3 + 3^3)^(1/2)`
