Advertisements
Advertisements
प्रश्न
Solve the following equation for x:
`4^(2x)=1/32`
Advertisements
उत्तर
`4^(2x)=1/32`
`rArr(2^2)^(2x)=1/2^5`
`rArr2^(4x)xx2^5=1`
`rArr2^(4x+5)=2^0`
⇒ 4x + 5 = 0
⇒ 4x = -5
`rArr x=-5/4`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
Find:-
`64^(1/2)`
Prove that:
`(x^a/x^b)^(a^2+ab+b^2)xx(x^b/x^c)^(b^2+bc+c^2)xx(x^c/x^a)^(c^2+ca+a^2)=1`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
Solve the following equation:
`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`
State the quotient law of exponents.
Write the value of \[\sqrt[3]{7} \times \sqrt[3]{49} .\]
If (23)2 = 4x, then 3x =
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
Simplify:
`11^(1/2)/11^(1/4)`
