Advertisements
Advertisements
प्रश्न
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
Advertisements
उत्तर
`3^(x-1)xx5^(2y-3)=225`
`rArr3^(x-1)xx5^(2y-3)=3xx3xx5xx5`
`rArr3^(x-1)xx5^(2y-3)=3^2xx5^2`
⇒ x - 1 = 2 and 2y - 3 = 2
⇒ x = 2 + 1 and 2y = 2 + 3
⇒ x = 3 and 2y = 5
⇒ x = 3 and y = 5/2
APPEARS IN
संबंधित प्रश्न
Simplify the following
`(a^(3n-9))^6/(a^(2n-4))`
Find the value of x in the following:
`(13)^(sqrtx)=4^4-3^4-6`
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.
Simplify \[\left[ \left\{ \left( 625 \right)^{- 1/2} \right\}^{- 1/4} \right]^2\]
The seventh root of x divided by the eighth root of x is
The square root of 64 divided by the cube root of 64 is
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
