Advertisements
Advertisements
Question
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
Advertisements
Solution
Given `(2^3)^4=(2^2)^x`
`2^(3xx4)=2^(2xx x)`
`2^12=2^(2x)`
On equating the exponents
12 = 2x
x = 12/2
x = 6
Hence, the value of x = 6.
APPEARS IN
RELATED QUESTIONS
Prove that:
`1/(1 + x^(b - a) + x^(c - a)) + 1/(1 + x^(a - b) + x^(c - b)) + 1/(1 + x^(b - c) + x^(a - c)) = 1`
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
If `5^(3x)=125` and `10^y=0.001,` find x and y.
If `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
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 10x = 64, what is the value of \[{10}^\frac{x}{2} + 1 ?\]
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
