Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

a.  64.4^x=4⁵

4³.4^x=4⁵

4^x=4⁵:4³

4^x=4²

x=2

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Lời giải:
a. $x^3=4^3\Rightarrow x=4$

b. $x^2=49=7^2=(-7)^2$

$\Rightarrow x=7$ hoặc $x=-7$

c. $x^3+1=28$

$x^3=28-1=27=3^3$

$\Rightarrow x=3$

d. $2^x=16=2^4$

$\Rightarrow x=4$

e. $2^4.2^x=2^6$

$\Rightarrow 2^{4+x}=2^6$

$\Rightarrow 4+x=6$

$\Rightarrow x=2$

g.

$5^x=25.5^3=5^2.5^3=5^5$

$\Rightarrow x=5$

Lần sau bạn lưu ý viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để đề được rõ ràng hơn nhé.

(x² + 1) + (x² + 2) + ... + (x² + 50) = 1475

x² + 1 + x² + 2 + ... + x² + 50 = 1475

50x² + (1 + 2 + ... + 50) = 1475

50x² + 50 . 51 : 2 = 1475

50x² + 1275 = 1475

50x² = 1475 - 1275

50x² = 200

x² = 200 : 50

x² = 4

x = 2 hoặc x = -2

Mình cho đề bài thế này nhé \(2^x+2^{x+1}+2^{x+2}+...+2^{x+2017}=2^{2020}-4\)             (1)

Nhân cả 2 vế của (1) cho 2, ta được \(2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2018}=2^{2021}-8\)             (2)

Lấy (2) trừ theo vế với (1), ta thu được \(2^{x+2018}-2^x=2^{2020}-4\)

\(\Leftrightarrow2^x.2^{2018}-2^x=2^2.2^{2018}-2^2.1\)

\(\Leftrightarrow2^x\left(2^{2018}-1\right)=2^2\left(2^{2018}-1\right)\)

do \(2^{2018}-1\ne0\) nên ta hoàn toàn có thể suy ra \(2^x=2^2\Leftrightarrow x=2\)

Vậy \(x=2\)

a) \(3^2.x+2^3.x=51\)

\(\Leftrightarrow x\left(3^2+2^3\right)=51\)

\(\Leftrightarrow17x=51\)

\(\Leftrightarrow x=3\)

Vậy

b) \(6^2.2-\left(84-3^2.x\right):7=69\)

\(\Leftrightarrow\left(84-3^2.x\right):7=3\)

\(\Leftrightarrow84-3^2.x=21\)

\(\Leftrightarrow3^2.x=63\)

\(\Leftrightarrow x=7\)

Vậy