Advertisements
Advertisements
प्रश्न
If `27^x=9/3^x,` find x.
Advertisements
उत्तर
We are given `27^x=9/3^x`
We have to find the value of x
Since `(3^3)^x=3^2/3^x`
By using the law of exponents `a^m/a^n=a^(m-n)` we get,
`3^(3x)=3^(2-x)`
on equating the exponents we get,
3x = 2 - x
3x + x = 2
4x = 2
x = 2/4
x = 1/2
Hence, `x=1/2`
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Simplify:
`root5((32)^-3)`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
If `3^(x+1)=9^(x-2),` find the value of `2^(1+x)`
Simplify:
`root(lm)(x^l/x^m)xxroot(mn)(x^m/x^n)xxroot(nl)(x^n/x^l)`
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
If o <y <x, which statement must be true?
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
