Advertisements
Advertisements
Question
Solve the following equation:
`3^(x+1)=27xx3^4`
Advertisements
Solution
`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
RELATED QUESTIONS
Simplify the following
`(a^(3n-9))^6/(a^(2n-4))`
Solve the following equations for x:
`2^(2x)-2^(x+3)+2^4=0`
If `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1),` prove that `a^(q-r)b^(r-p)c^(p-q)=1`
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
If ax = by = cz and b2 = ac, show that `y=(2zx)/(z+x)`
Write the value of \[\sqrt[3]{7} \times \sqrt[3]{49} .\]
If \[2^{- m} \times \frac{1}{2^m} = \frac{1}{4},\] then \[\frac{1}{14}\left\{ ( 4^m )^{1/2} + \left( \frac{1}{5^m} \right)^{- 1} \right\}\] is equal to
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 =
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
