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
If a = 3 and b = -2, find the values of :
ab + ba
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Find the value of x in the following:
`5^(x-2)xx3^(2x-3)=135`
If a and b are different positive primes such that
`(a+b)^-1(a^-1+b^-1)=a^xb^y,` find x + y + 2.
If 1176 = `2^axx3^bxx7^c,` find the values of a, b and c. Hence, compute the value of `2^axx3^bxx7^-c` as a fraction.
Show that:
`((a+1/b)^mxx(a-1/b)^n)/((b+1/a)^mxx(b-1/a)^n)=(a/b)^(m+n)`
If (x − 1)3 = 8, What is the value of (x + 1)2 ?
If \[\frac{5 - \sqrt{3}}{2 + \sqrt{3}} = x + y\sqrt{3}\] , then
