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\(4^3\cdot4^{x-1}=64\)
\(\Leftrightarrow4^{x-1}=1\)
\(\Leftrightarrow x-1=0\)
hay x=1
a: 3x=81
nên x=27
b: \(5\cdot4^x=80\)
\(\Leftrightarrow4^x=16\)
hay x=2
c: \(2^x=4^5:4^3\)
\(\Leftrightarrow2^x=2^4\)
hay x=4
`64.4^x=168`
`<=>4^x=168/64=21/8`
Vì `x\inNN` nên không tồn tại `x`
Kết luận:.....
\(318-5\left(x-64\right)=103\)
\(\Rightarrow5\left(x-64\right)=318-103\)
\(\Rightarrow5\left(x-64\right)=215\)
\(\Rightarrow x-64=43\)
\(\Rightarrow x=43+64\)
\(\Rightarrow x=107\)
_____________
\(4^x\cdot5+216=296\)
\(\Rightarrow4^x\cdot5=296-216\)
\(\Rightarrow4^x\cdot5-80\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
___________
\(376-6^x:3=364\)
\(\Rightarrow6^x:3=376-364\)
\(\Rightarrow6^x:3=12\)
\(\Rightarrow6^x=36\)
\(\Rightarrow6^x=6^2\)
\(\Rightarrow x=2\)
___________
\(\left(4x-1\right)^2=121\)
\(\Rightarrow\left(4x-1\right)^2=11^2\)
\(\Rightarrow\left[{}\begin{matrix}4x-1=11\\4x-1=-11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=12\\4x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
c) \(\Rightarrow2^x=32\Rightarrow x=5\)
d) \(\Rightarrow4^3.4^x=4^5\Rightarrow4^x=4^2\Rightarrow x=2\)
e) \(\Rightarrow3^3.3^x=3^5\Rightarrow3^x=3^2\Rightarrow x=2\)
f) \(\Rightarrow7^2.7^x=7^4\Rightarrow7^x=7^2\Rightarrow x=2\)
a. 2x . 4 = 128
<=> 2x + 2 = 27
<=> x + 2 = 7
<=> x = 5
b. (2x + 1)3 = 125
<=> (2x + 1)3 - 53 = 0
<=> (2x + 1 - 5)\(\left[\left(2x+1\right)^2+\left(2x+1\right).5+25\right]=0\)
<=> (2x - 4)(4x2 + 4x + 1 + 10x + 5 + 25) = 0
<=> (2x - 4)(4x2 + 14x + 31) = 0
<=> \(\left[{}\begin{matrix}2x-4=0\\4x^2+14x+31=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\VôNghiệm\end{matrix}\right.\)
c. 2x - 26 = 6
<=> 2x = 32
<=> x = 5
d. 64 . 4x = 45
<=> 43 . 4x = 45
<=> 43 + x = 45
<=> 3 + x = 5
<=> x = 2
e. 27 . 3x = 243
<=> 33 . 3x = 35
<=> 33 + x = 35
<=> 3 + x = 5
<=> x = 2
g. 49 . 7x = 2401 (Bn xem lại đề câu này)
<=> 72 . 7x = 74
<=> 72 + x = 74
<=> 2 + x = 4
<=> x = 2
(64-4x).3=24
64-4x=24:3
64-4x=8
4x=64-8
4x=56
x=56:4
x=14
vậy x=14
Ta có (4x-12)(x³+64)=0
Suy ra 4x - 12 = 0 hoặc x³ + 64 = 0
=> 4x = 12 hoặc x³ = - 64
=> x = 3 hoặc x = - 4
Vậy x = 3 hoặc x = - 4
(4x-12)(x³+64)=0
\(\Rightarrow\orbr{\begin{cases}4x-12=0\\x^3+64=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}4x=12\\x^3=-64\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=3\\x^3=\left(-4\right)^3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
Vậy \(x\in\left\{3;-4\right\}\)
Hok tốt !