Advertisements
Advertisements
प्रश्न
Solve the following equation:
`3^(x+1)=27xx3^4`
उत्तर
`3^(x+1)=27xx3^4`
`rArr3^(x+1)=3^3xx3^4`
`rArr3^(x+1)=3^(3+4)`
`rArr3^(x+1)=3^7`
⇒ x + 1 = 7
⇒ x = 7 - 1
⇒ x = 6
APPEARS IN
संबंधित प्रश्न
Find:-
`64^(1/2)`
Simplify the following
`3(a^4b^3)^10xx5(a^2b^2)^3`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Assuming that x, y, z are positive real numbers, simplify the following:
`root5(243x^10y^5z^10)`
Prove that:
`9^(3/2)-3xx5^0-(1/81)^(-1/2)=15`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
Show that:
`((a+1/b)^mxx(a-1/b)^n)/((b+1/a)^mxx(b-1/a)^n)=(a/b)^(m+n)`
Write the value of \[\sqrt[3]{125 \times 27}\].
If (x − 1)3 = 8, What is the value of (x + 1)2 ?
If x = \[\frac{2}{3 + \sqrt{7}}\],then (x−3)2 =
