Advertisements
Advertisements
प्रश्न
If 1176 = `2^axx3^bxx7^c,` find the values of a, b and c. Hence, compute the value of `2^axx3^bxx7^-c` as a fraction.
Advertisements
उत्तर
First find the prime factorisation of 1176.
It can be observed that 1176 can be written as `2^3xx3^1xx7^2`
`1176=2^a3^b7^c=2^3 3^1 7^2`
So, a = 3, b = 1 and c = 2.
Therefore, the value of `2^a xx3^bxx76-c` is `2^3xx3^1xx7^-2=8xx3xx1/49=24/49`
APPEARS IN
संबंधित प्रश्न
Solve the following equation for x:
`4^(2x)=1/32`
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^-3))^5`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt2/sqrt3)^5(6/7)^2`
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
State the quotient law of exponents.
If \[\frac{3^{2x - 8}}{225} = \frac{5^3}{5^x},\] then x =
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
