Advertisements
Advertisements
Question
If \[\frac{3^{2x - 8}}{225} = \frac{5^3}{5^x},\] then x =
Options
2
3
5
4
Advertisements
Solution
We have to find the value of x provided \[\frac{3^{2x - 8}}{225} = \frac{5^3}{5^x},\]
So,
\[\frac{3^{2x - 8}}{3^2 × 5^2} = \frac{5^3}{5^x}\]
By cross multiplication we get
`3^(2x-8) xx 5^x = 3^2xx5^2 xx5^3`
By equating exponents we get
`3^(2x-8) = 3^2`
`2x - 8 = 2`
`2x= 2+8`
`2x = 10`
`x=10/2`
`x=5`
And
`5^x = 5^(3+2)`
`x=3+2`
`x=5`
APPEARS IN
RELATED QUESTIONS
Prove that:
`(a+b+c)/(a^-1b^-1+b^-1c^-1+c^-1a^-1)=abc`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Simplify:
`(16^(-1/5))^(5/2)`
Prove that:
`(1/4)^-2-3xx8^(2/3)xx4^0+(9/16)^(-1/2)=16/3`
Find the value of x in the following:
`2^(x-7)xx5^(x-4)=1250`
If `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
`(2/3)^x (3/2)^(2x)=81/16 `then x =
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
Find:-
`32^(2/5)`
Which of the following is equal to x?
