Advertisements
Advertisements
प्रश्न
Solve the following equation for x:
`2^(3x-7)=256`
Advertisements
उत्तर
`2^(3x-7)=256`
`rArr2^(3x-7)=2^8`
⇒ 3x - 7 = 8
⇒ 3x = 8 + 7
⇒ 3x = 15
⇒ x = 15/3
⇒ x = 5
APPEARS IN
संबंधित प्रश्न
Simplify the following
`(a^(3n-9))^6/(a^(2n-4))`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Show that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
Find the value of x in the following:
`5^(x-2)xx3^(2x-3)=135`
If `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
If \[8^{x + 1}\] = 64 , what is the value of \[3^{2x + 1}\] ?
Which one of the following is not equal to \[\left( \frac{100}{9} \right)^{- 3/2}\]?
The value of \[\left\{ \left( 23 + 2^2 \right)^{2/3} + (140 - 19 )^{1/2} \right\}^2 ,\] is
If x= \[\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\] and y = \[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\] , then x2 + y +y2 =
Simplify:
`(9^(1/3) xx 27^(-1/2))/(3^(1/6) xx 3^(- 2/3))`
