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a) \(\left(-\frac{1}{2}\right)^n=\frac{1}{64}\)
\(\left(-\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^6\)
\(\left(-\frac{1}{2}\right)^n=\left(-\frac{1}{2}\right)^6\)
\(\Rightarrow n=6\)
b) \(\frac{2^n}{32}=2\)
\(2^n=2.32\)
\(2^n=64\)
\(2^n=2^6\)
\(\Rightarrow n=6\)
c) \(\left(-3\right)^3.\left(-3\right)^n=-243\)
\(\left(-3\right)^{3+n}=\left(-3\right)^5\)
\(\Rightarrow3+n=5\)
\(n=5-3\)
\(n=2\)
a, ( - 1/2)n = (1/2)6
n = 6
2n = 2 . 32 = 64
2n = 26
n = 6
(-3)3 . (-3)n = (-3)5
(-3)n = (-3)2
n = 2
Bài 10:
a) (1/3)n = 1/81
=> (1/3)n = (1/3)4
=> n = 4
b) -512/343 = (-8/7)n
=> (-8/7)3 = (-8/7)n
=> 3 = n (hay n = 3)
c) (-3/4)n = 81/256
=> (-3/4)n = (-3/4)4
=> n = 4
d) 64/(-2)n = (-2)3
=> 64/(-2)n = -8
=> (-2)n = -8
=> (-2)n = (-2)3
=> n = 3
Bài 11: (không có y để tìm nhé)
a) (0,4x - 1,3)2 = 5,29
=> (0,4x - 1,3)2 = (2,3)2
=> 0,4x - 1,3 = 2,3
=> 0,4x = 3,6
=> x = 9
b) (3/5 - 2/3x)3 = -64/125
=> (3/5 - 2/3x)3 = (-4/5)3
=> 3/5 - 2/3x = -4/5
=> 2/3x = 7/5
=> x = 21/10
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{27}\)
\(\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow n=3\)
b) \(\left(\frac{3}{5}\right)^n=\frac{81}{625}\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^4\)
\(\Rightarrow n=4\)
c) \(3^n\cdot2^n=36\)
\(\left(3\cdot2\right)^n=36\)
\(6^n=6^2\)
\(\Rightarrow n=2\)
d) \(\frac{2^n}{3^n}=\frac{8}{27}\)
\(\left(\frac{2}{3}\right)^n=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow n=3\)
1.
a) 2n : 4 = 16
2n = 16.4
2n = 64
2n = 26
=> n = 6
b) 5n = 625
5n = 54
=> n = 4
c) 2n . 16 = 128
2n = 128 : 16
2n = 8
2n = 23
=> n = 3
d) 3n+3 = 81
3n+3 = 34
=> n + 3 = 4
=> n = 4 - 3
=> n = 1
1.
a) 2n : 4 = 16
2n = 16.4
2n = 64
2n = 26
=> n = 6
b) 5n = 625
5n = 54
=> n = 4
c) 2n . 16 = 128
2n = 128 : 16
2n = 8
2n = 23
=> n = 3
d) 3n+3 = 81
3n+3 = 34
=> n + 3 = 4
=> n = 4 - 3
=> n = 1
2.
a) x3 = 64
x3 = 43
=> x = 4
b) (2x + 1)3 = 27
(2x + 1)3 = 33
2x + 1 = 3
2x = 3 - 1
2x = 2
=> x = 1
c) x3 = x
=> x = {0;1}
d) 25 + 5x = 75.73
25 + 5x = 5764801
5x = 5764801 - 25
5x = 5764776
=> không có số tự nhiên x thoản mãn
3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
\(\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{27}\right)\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow n=3\)
\(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)
\(\Rightarrow n=4\)
a, \(\left(\dfrac{1}{3}\right)^n=\dfrac{1}{27}\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)
Vì \(\dfrac{1}{3}\ne-1,\dfrac{1}{3}\ne0;\dfrac{1}{3}\ne1\) nên \(n=3\)
Vậy........
b, \(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)
Vì \(\dfrac{3}{5}\ne-1,\dfrac{3}{5}\ne0;\dfrac{3}{5}\ne1\) nên \(n=4\)
Vậy..........
Chúc bạn học tốt!!!