tìm stn bt 4 mũ n = 64.4
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\(2^n-64=26\Rightarrow2^n=90\Rightarrow n=log_290\notinℕ\)Đề lag r
64 . 4x = 168
<=> 43. 4x = 416
=> 3 + x = 16
<=> x = 13
Vậy x = 13
2x.162 = 1024
<=> 2x. 28 = 210
=> x + 8 = 10
<=> x = 2
Vậy x = 2
b: Ta có: \(2^x\cdot16^2=1024\)
\(\Leftrightarrow2^x\cdot2^8=2^{10}\)
\(\Leftrightarrow x+8=10\)
hay x=2
A = 4+22+23+...+299
2A = 23+23+24+...+2100
2A - A = 23 +(2100 - 23)
=> A = 2100
Có A.214 = 2n
=> 2100.214 = 2n
=> 2114 = 2n
=> n = 114
Lời giải:
$A=5^{50}-5^{48}+5^{46}-5^{44}+....-5^4+5^2-1$
$5^2A=5^{52}-5^{50}+5^{48}-5^{46}+...-5^6+5^4-5^2$
$\Rightarrow A+5^2A=5^{52}-1$
$\Rightarrow 26A=5^{52}-1$
$\Rightarrow 5^{52}-1+1=5^n$
$\Rightarrow 5^{52}=5^n$
$\Rightarrow n=52$
a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)
b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)
c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)
d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)
f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)
g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)
h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)
i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4n = 4096
4n = 212
n = 12
5n = 15625
5n = 56
n = 6
6n+3 = 216
6n+3 = 23.33
6n+3 = 63
n + 3 = 3
+/\(2^n=32\)(=) \(2^n=2^5\)
=> \(n=5\)
+/\(64.4^n=4^5\) (=) \(4^3.4^n=4^5\)
(=)\(4^n=4^2\) => \(n=2\)
Các ý còn lại bạn tự làm nhé !!
2^5 = 32
64*4^2=4^5
27*3^2=243
49*7^2=2401
9<3^3<81
Đây là tìm n nhé bn
Ta có: \(64\cdot4^x=168\)
\(\Leftrightarrow4^x=\dfrac{21}{8}\)
hay \(x\in\varnothing\)
Ta có:
\(A=3+3^2+3^3+...+3^{100}\)
=> \(3A=3^2+3^3+3^4+...+3^{101}\)
=> \(3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)
<=> \(2A=3^{101}-3\)
Thay vào PT ta được: \(2A+3=3^n\)
\(\Rightarrow3^n=3^{101}-3+3=3^{101}\)
\(\Rightarrow n=101\)
Bài làm :
\(4^n=64.4\)
\(\Rightarrow4^n=256\)
\(\Rightarrow4^n=4^4\)
\(\Rightarrow n=4\)
Vậy n = 4 .
Học tốt
\(4^n=64.4\)
\(4^n=256\)
\(4^n=4^4\)
\(n=4\)
vậy \(n=4\)