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a: 3^x=243
=>3^x=3^5
=>x=5
b: 4^x=4096
=>4^x=4^5
=>x=5
c: 5^3-x=25
=>3-x=2
=>x=1
d: =>2x-3=3
=>2x=6
=>x=3
j: =>2^x*8+2^x*2=80
=>2^x=8
=>x=3
42n+2 = 4069 => 42n+2 = 46 => 2n + 2 = 6 => 2n = 6 - 2 => 2n = 4 => n = 4 : 2 => n = 2
A=1+2+2^2+...+2^13
=>2A=2+2^2+...+2^14
=>2A-A=2^14-1
=>A=2^14-1
(x+6)4=4096
(x+6)4=84
==> x+6=8 hoặc x+6=—8
==> x=8–6 hoặc x=—8–6
==> x= 2 hoặc x=—14
2x—3=128
2x—3=27
==> x—3=7
x=7+3
x=10
Ss: 22018 và 16900
Ta có 16900=(24)900=23600
Vì 22018<23600
Nên 22018<23600
A = 1+2+4+8+...+8192
A = 2 0 + 2 1 + 2 2 + 2 3 +...+2 13
2A = 2 1+2 2+2 3+2 4+...+2 14
2A ‐ A = 2 14 ‐ 2 0
=> A = 2 14 ‐ 1
a,2^x*16=128
2^x=8
2^x=2^3
suy ra x=3
b,3^x/9=27
3^x/3^2=3^3
3^x=3^5
suy ra x=5
c,(2*x+1)^3=27
(2*x+1)^3=3^3
suy ra 2*x+1=3
suy ra 2x=2
suy ra x=1
d,(x*2)^2=(x*2)^4
suy ra (x*2)^4-(x*2)^2=0
(x*2)^2*((x*2)^2-1)=0
suy ra (x*2)^2=0 hoac (x*2)^2-1=0
+(x*2)^2=0 +(x*2)^2-1=0
suy ra x*2=0 (x*2)^2=1
suy ra x=0 suy ra x*2=1 hoac -1
suy ra x=1/2 hoac -1/2
= 16781312 + 1 = 16781313
k tớ nhé
k ết bạn nha
li -ke cho tớ thật nhiều tuần sau li -ke lại cho
Đặt tổng trên = A
\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{4096}+\frac{1}{8192}\)
\(A.2=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}+\frac{1}{4096}\)
\(A.2-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}+\frac{1}{4096}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{4096}+\frac{1}{8192}\right)\)
\(A=2-\frac{1}{8192}=\frac{16383}{8192}\)
Đặt A = 1 + 1/2 + 1/4 + 1/8 + ... + 1/4096 + 1/8192
2A = 2 + 1 + 1/2 + 1/4 + ... + 1/2048 + 1/4096
2A - A = (2 + 1 + 1/2 + 1/4 + ... + 1/2048 + 1/4096) - (1 + 1/2 + 1/4 + 1/8 +... + 1/4096 + 1/8192)
A = 2 - 1/8192
A = 16383/8192
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
Ta có (x-2)^4=4096
suy ra:x-2=8
x=8+2
x=10
( x - 2 ) \(^4\)=4096
=> ( x-2 ) \(^4\)=2\(^{12}\)
=> ( x-2 ) \(^4\)=( 2\(^3\))\(^4\)
=> ( x - 2 ) \(^4\)= 8\(^4\)
=> x - 2 = 8
x = 8+2
x = 10