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(2/3 )^3 x (2/3)^5 = 2/3^ n-2
(2/3)^3+5 = 2/3 ^ n-2
2/3 ^ 8= 2/3 ^n-2
now 2/3 are canceled with each other
8= n-2
8+ 2 =n
n= 10
(2/3)^3+5 = 2/3 ^ n-2
2/3 ^ 8= 2/3 ^n-2
now 2/3 are canceled with each other
8= n-2
8+ 2 =n
n= 10
Answered by
0
Hello friend
_____________________________________
(2/3)³×(2/3)^5=(2/3)^n-2
a^m × a^n = a^(m+n)
(2/3)³×(2/3)^5=(2/3)^3+5
(2/3)^8=(2/3)^n-2
Equating the powers as the bases are equal
8=n-2
n=8+2
n=10
_____________________________________
-3^n+1 × -3^5 = -3^-4
a^m × a^n = a^(m+n)
-3^(n+1+5) = -3^-4
As the bases are equal we can equate the powers
n+6= -4
n= -4-6 = -10
_____________________________________
7^2n+1 ÷ 49 = 7³
49=7²
7^2n+1 ÷ 7² = 7³
7^2n+1 = 7³ × 7²
a^m × a^n = a^(m+n)
7^2n+1 = 7^5
As the bases are same we can equate the powers
2n+1=5
n=4/2=2
_____________________________________
Hope it helps!
_____________________________________
(2/3)³×(2/3)^5=(2/3)^n-2
a^m × a^n = a^(m+n)
(2/3)³×(2/3)^5=(2/3)^3+5
(2/3)^8=(2/3)^n-2
Equating the powers as the bases are equal
8=n-2
n=8+2
n=10
_____________________________________
-3^n+1 × -3^5 = -3^-4
a^m × a^n = a^(m+n)
-3^(n+1+5) = -3^-4
As the bases are equal we can equate the powers
n+6= -4
n= -4-6 = -10
_____________________________________
7^2n+1 ÷ 49 = 7³
49=7²
7^2n+1 ÷ 7² = 7³
7^2n+1 = 7³ × 7²
a^m × a^n = a^(m+n)
7^2n+1 = 7^5
As the bases are same we can equate the powers
2n+1=5
n=4/2=2
_____________________________________
Hope it helps!
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