Tìm X biết:
2x+2 = 2401
Câu 4 : Tìm x, biết.
a) 2^x.4 = 128 b) (2x + 1)^3 = 125 c) 2x – 2^6 = 6 d) 49.7^x = 2401\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)
Tìm x biết
(2x+3)^4=2401
\(\left(2x+3\right)^4=2401\)
\(\Rightarrow\left(2x+3\right)^4=7^4\)
=> 2x - 3 = 7 hoặc 2x - 3 = -7
x = 5 x = -2
\(\left(2x+3\right)^4=2401\)
\(\left(2x+3\right)^4=7^4\)
\(\Rightarrow\orbr{\begin{cases}2x+3=7\\2x+3=-7\end{cases}\Rightarrow\orbr{\begin{cases}2x=7-3=4\\2x=\left(-7\right)-3=-10\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
Vậy ____
Câu 3: Tìm x ∈ N, biết:
a) 3 x . 3 = 243 b) 2 x . 162 = 1024 c) 64.4x = 168 d) 2 x = 16Câu 4 : Tìm x, biết. a) 2 x .4 = 128 b) (2x + 1)3 = 125 c) 2x – 2 6 = 6 d) 49.7x = 24013:
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
Tìm x ∈ N, biết.
a, 2 x . 2 2 = 32
b, 27 . 3 x = 243
c, 2 x . 2 4 = 1024
d, 49 . 7 x = 2401
a, 2 x . 2 2 = 32
2 x + 2 = 2 5
x + 2 = 5
x = 3
Vậy x = 3
b, 27 . 3 x = 243
3 3 . 3 x = 3 5
3 3 + x = 3 5
x + 3 = 5
x = 2
Vậy x = 2
c, 2 x . 2 4 = 1024
2 x + 4 = 2 10
x + 4 = 10
x = 6
Vậy x = 6
d, 49 . 7 x = 2401
7 2 . 7 x = 7 4
7 2 + x = 7 4
2 + x = 4
x = 2
Vậy x = 2
Tìm x biết
( 3 x X - 16 ) x 343 - 2 x 2401
7^2x+1:7^x=2401
72x+1-x=74
7x+1=74
=>x+1=4
x=4-1
x=3
Vậy x=3
(x-1 phần 2) mũ 2 =0
(x-2) mũ 2=1
(2x-1) mũ 3 =-8
(1 phần 2) mũ 2=1phần 16
49 nhân 7 mũ x =2401
2x mũ x - 15=17
2 mũ x+3 nhân 2 mũ x =32
2x (x-1 phần 7)=0
tìm x hộ mình nhé
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
\(49\times7^x=2401\)
\(7^x=2401:49\)
\(7^x=49\)
\(7^x=7^2\)
\(\Rightarrow x=2\)
\(2x^x-15=17\)
\(2x^x=17+15\)
\(2x^x=32\)
\(2x^x=2^5\)
\(\Rightarrow x\times x=5\)
\(\Rightarrow x^2=5\)
\(2^x+3\times2^x=32\)
\(2^x\times\left(3+1\right)=32\)
\(2^x\times4=32\)
\(2^x=32:4\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
\(2x\left(x-\frac{1}{7}\right)=0\)
\(x\left(x-\frac{1}{7}\right)=0:2\)
\(x\left(x-\frac{1}{7}\right)=0\)
\(x\times x-x\times\frac{1}{7}=0\)
\(x^2-x\frac{1}{7}=0\)
\(x-x=0:\frac{1}{7}\)
\(x-x=0\)
\(\Rightarrow x=0\)
Tìm số tự nhiên \(n\), biết \(n^4=2401\)
n4 = 2401
(n2)2 = (72)2
n2 = 72
\(\left[{}\begin{matrix}n=7\\n=-7\end{matrix}\right.\)
vì n ϵ N nên n = 7
Kết luận n = 7 là giá trị thỏa mãn yêu cầu đề bài.
72 + 72x+2 = 2401
7^2+7^2x+2=2401
=>7^2+7^2x.7^2=2401
=>7^2.(1+7^2x)=2401
=>1+7^2x+1=49=7^2
=>....