`#3107.101107`

a)

\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)

Vậy, `x = 4`

b)

\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)

Vậy, `x = 3`

c)

\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)

Vậy, `x = 4.`

a) \(27< 3^x< 243\)

\(\Rightarrow3^3< 3^x< 3^5\)

\(\Rightarrow3< x< 5\)

c) \(3^x+3^{x+2}=810\)

\(\Rightarrow3^x\left(1+3^2\right)=810\)

\(\Rightarrow3^x.10=810\)

\(\Rightarrow3^x=810:10\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

\(3^{2x+1}=243\)

\(\Rightarrow3^{2x+1}=3^5\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=5-1\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

\(3^2x+1=243\)

\(9x=243-1\)

\(9x=242\)

\(x=242:9\)

\(\Rightarrow x=\dfrac{242}{9}\)

A) \(3^x\cdot3=243\)

\(\Leftrightarrow3^x=243:3=81\)

\(\Leftrightarrow3^x=3^4\)

\(\Leftrightarrow x=4\)

B) \(2^x-15=17\)

\(\Leftrightarrow2^x=17+15\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

C) \(\left(x-1\right)^2-7=42\)

\(\Leftrightarrow\left(x-1\right)^2=49\)

\(\Rightarrow\orbr{\begin{cases}x-1=7\\x-1=-7\end{cases}\Rightarrow\orbr{\begin{cases}x=8\\x=-6\end{cases}}}\)

Câu a thì có nhiều cách nhưng mình chỉ làm 1 cách thôi nhé

a) 3^x . 3 = 243

=> 3^x . 3¹ = 3⁵

3^x = 3⁵ : 3¹

3^x = 3^5-1

3^x = 3⁴

<=> x = 4

b) 2^x - 15 = 17

2^x = 17 - 15

2^x = 2

<=> 2^x = 2¹ => x = 1

c) (x-1)² - 7 = 42

(x-1)² = 42 + 7

(x-1)² = 49

x-1 = 49 = 7.7

<=> x-1 = 7

x = 7 + 1

x = 8

(3^x)² : 3³ = \(\frac{1}{243}\)

<=> 3^2x : 3³ = \(\frac{1}{3^5}\)

<=> 3^{2x - 3} = 3^-5

=> 2x - 3 = -5

     2x      = -5 + 3

     2x      = -2

       x      = -2 : 2

       x       = -1

\(\left(3^x\right)^2:3^3=\frac{1}{243}\Rightarrow3^{2x}:3^3=\frac{1}{3^5}\Rightarrow3^{2x-3}=3^{-5}\Rightarrow2x-3=-5\Rightarrow2x=-2\Rightarrow x=-1\)

a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

=>\(x< -\dfrac{3}{2}\)

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

\(3^{x+1}-2.3^x=243\\ \Rightarrow3^x.3-2.3^x=243\\ \Rightarrow3^x=3^5\\ \Rightarrow x=5\)

\(d.3^{x+1}-2.3^x=243\)