a) \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

\(\Rightarrow\frac{1}{3^m}=\frac{1}{81}\)

<=> 3^m = 81

=> 3^m = 3⁴ ( 81 = 3⁴ )

<=> m = 4

b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\left(\frac{3}{5}\right)^n=\frac{9}{9765625}\)

\(\Rightarrow\frac{3}{5^n}=\frac{9}{9765625}\)

=> 5ⁿ = 9765625

=> 5ⁿ = 5¹⁰ ( 9765625 = 5¹⁰ )

<=> n = 10

\(\left(-0,25\right)^p=\frac{1}{256}\)

\(\left(\frac{-1}{4}\right)^p=\frac{1}{256}\)

\(\Rightarrow\frac{-1}{4^p}=\frac{1}{256}\)

=> 4^p = 256

=> 4^p = 4⁴ ( 256 = 4⁴ )

<=> p = 4

a ) \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow m=4\)

b ) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

       \(\Leftrightarrow\left(\frac{3}{5}^2\right)^n=\left(\frac{9}{25}\right)^5\)

       \(\Leftrightarrow\left(\frac{9}{25}\right)^n=\left(\frac{9}{25}\right)^5\)

       \(\Leftrightarrow n=5\)

c ) \(\left(-0,25\right)^p=\frac{1}{256}\)

   \(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)

   \(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\left(-\frac{1}{4}\right)^4\)

   \(\Leftrightarrow p=4\)

\(a.\)

\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

\(\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow m=4\)

Vậy :        \(m=4\)

\(b.\)

\(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{15}\)

\(\Rightarrow n=5\)

Vậy :        \(n=5\)

\(c.\)

\(\left(-0,25\right)^p=\frac{1}{256}\)

\(\Rightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)

\(\Rightarrow\left(-\frac{1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)

\(\Rightarrow p=4\)

Vậy :        \(p=4\)

a) \(\left(\dfrac{1}{3}\right)^m=\dfrac{1}{81}\)

\(\Rightarrow\dfrac{1^m}{3^m}=\dfrac{1}{81}\)

\(\Rightarrow\dfrac{1}{3^m}=\dfrac{1}{3^4}\)

\(\Rightarrow m=4\)

b) \(\left(\dfrac{3}{5}\right)^n=\left(\dfrac{9}{25}\right)^5\)

\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left[\left(\dfrac{3}{5}\right)^2\right]^5\)

\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^{10}\)

\(\Rightarrow n=10\)

c) \(\left(-0,25\right)^p=\dfrac{1}{256}\)

\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\dfrac{1}{256}\)

\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\dfrac{1}{4^4}\)

\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\left(\dfrac{1}{4}\right)^4\)

\(\Rightarrow p=4\)

a/ \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

\(\Leftrightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Leftrightarrow m=4\left(tm\right)\)

b/ \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\Leftrightarrow n=10\)

\(\Leftrightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)

a)(1/3)^m=(1/3)^4

b)(3/5)^n=(3/5)^10

c)(-0,25)^p=(-0,25)^4

a)\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

 \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow m=4\)

vay \(m=4\)

b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)

\(\Rightarrow n=10\)

vay \(n=10\)

c) \(\left(-0,25\right)^p=\frac{1}{256}\)

\(\left(\frac{-1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)

\(\Rightarrow p=4\)

voi mu duong thi \(\left(\frac{-1}{4}\right)^4=\left(\frac{1}{4}\right)^4\)

vay \(p=4\)

(1/3)^4=1/81

(3/5^10)=(9/25)

(-0,25)^4=1/256

Bài 1

a) x:(3/4)³=(3/4)²=>x=(3/4)².(3/4)³=(3/4)²⁺³=(3/4)⁵

b) (2/5)⁵.x=(2/5)⁸=>x=(2/5)⁸:(2/5)⁵=(2/5)^8-5=(2/5)³

Bài 2

(0,36)⁸=[ (0,6)² ]⁸=(0,6)^2.8=(0,6)¹⁶

(0,216)⁴=[ (0,6)³ ]⁴=(0,6)^3.4=(0,6)¹²

Bài 3

a) (1/3)^m=1/81=(1/3)⁴ => m=4

b) (3/5)ⁿ=(9/25)⁵=[ (3/5)² ]⁵=(3/5)^2.5=(3/5)¹⁰=>n=10

c) (-0,25)^p=1/256=(1/4)⁴=(0,25)⁴=(-0,25)⁴=>p=4

bai3

a (1/3)^m=(1/3)^4 =>m=4

b) n=10 (làm tuong tu cau a)

c ) p=4 (doi -0.25=-1/4)

cau 2(0.36)^8=0.6)^16

(0.216)^4=0.6)12

Tính chất bạn cho vẫn thiếu trong quá trình làm bài mình sẽ bổ sung!

a, \(\left(\dfrac{1}{3}\right)^m=\dfrac{1}{81}\)

\(\Rightarrow\left(\dfrac{1}{3}\right)^m=\left(\dfrac{1}{3}\right)^4\)

Vì \(\dfrac{1}{3}\ne\pm1;\dfrac{1}{3}\ne0\) nên \(m=4\)

b, \(\left(\dfrac{3}{5}\right)^n=\left(\dfrac{9}{25}\right)^5\)

\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^{10}\)

Vì \(\dfrac{3}{5}\ne\pm1;\dfrac{3}{5}\ne0\) nên \(n=10\)

a)\(\left(\dfrac{1}{3}\right)^m=\dfrac{1}{81}\)

=>\(\left(\dfrac{1}{3}\right)^m=\left(\dfrac{1}{3}\right)^4\)

=>\(m=4\)

b)\(\left(\dfrac{3}{5}\right)^m=\left(\dfrac{9}{25}\right)^5\)

=>\(\left(\dfrac{3}{5}\right)^m=\left(\dfrac{3}{5}\right)^{10}\)

=>\(m=10\)

c)\(\left(-0,25\right)^p=\dfrac{1}{256}\)

=>\(\left(-0,25\right)^p=\left(\dfrac{1}{4}\right)^4\)

=>\(p=4\)