Advertisements
Advertisements
Question
Solve the following equation for x:
`4^(2x)=1/32`
Solution
`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
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Simplify:-
`2^(2/3). 2^(1/5)`
Simplify the following
`((x^2y^2)/(a^2b^3))^n`
Solve the following equation for x:
`2^(5x+3)=8^(x+3)`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Find the value of x in the following:
`5^(2x+3)=1`
If \[8^{x + 1}\] = 64 , what is the value of \[3^{2x + 1}\] ?
The value of \[\left\{ \left( 23 + 2^2 \right)^{2/3} + (140 - 19 )^{1/2} \right\}^2 ,\] is
If o <y <x, which statement must be true?
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=
Simplify:-
`(1/3^3)^7`
