Advertisements
Advertisements
प्रश्न
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Advertisements
उत्तर
`3^(2x+4)+1=2.3^(x+2)`
`rArr(3^(x+2))^2-2.3^(x+2)+1=0`
`rArr(3^(x+2)-1)^2=0`
`rArr3^(x+2)-1=0`
`rArr3^(x+2)=1`
`rArr3^(x+2)=3^0`
⇒ x + 2 = 0
⇒ x = -2
APPEARS IN
संबंधित प्रश्न
Simplify the following
`((x^2y^2)/(a^2b^3))^n`
If a = 3 and b = -2, find the values of :
aa + bb
Solve the following equation for x:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Simplify:
`root3((343)^-2)`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
If `a=x^(m+n)y^l, b=x^(n+l)y^m` and `c=x^(l+m)y^n,` Prove that `a^(m-n)b^(n-l)c^(l-m)=1`
If x-2 = 64, then x1/3+x0 =
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
