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a) (2^x).4=128
2^x = 128:4
2^x = 32
mà 32=2^5=>x=5
b) ta có: x^15=x
theo quy ước: 0^15=0;1^15=1
=> x=1
4 câu còn lại mai mình sẽ giải nhé
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a) \(2^x.4=128\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
vậy \(x=5\)
b) \(x^{15}=x\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
c) \(2^x.\left(2^2\right)^2=\left(2^3\right)^2\)
\(2^x.2^4=2^6\)
\(2^x=2^2\)
\(\Rightarrow x=2\)
vậy \(x=2\)
d) \(\left(x^5\right)^{10}=x\)
\(x^{50}=x\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
a, 2x.4 = 128
2x = 128 : 4
2x = 32
2x = 25
=> x = 5
b, x15 = x
=>x15 - x = 0
=> x14.x - x = 0
=> x(x14 - 1) = 0
=> x = 0 hoặc x14 - 1 =0
=> x = 0 hoặc x14 = 1
=> x = 0 hoặc x = 1 hoặc x = -1
Mà x \(\in\)N
=> x = 0 hoặc x = 1
c, 2x.(22)2 = (23)2
2x.24 = 26
2x = 26 : 24
2x = 22
=> x = 2
d, (x5)10 = x
=> x50 = x
=> x50 - x = 0
=> x49.x - x = 0
=> x(x49 - 1) = 0
=> x = 0 hoặc x49 - 1 = 0
=> x = 0 hoặc x49 = 1
=> x = 0 hoặc x = 1
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\(2^x.4=128\)
\(\Rightarrow2^x=32\left(\text{cùng chia cho 4}\right)\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
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\(a,2^x\cdot4=128\)
\(2^x=128:4=32\)
\(2^x=2^5\)
\(x=5\)
\(b,x^{15}=x\)
\(x^{15}-x=0\)
\(x\left(x^{14}-1\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x^{14}-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(c,16^x< 128\)
\(2^{4x}< 2^7\)
\(4x< 7\)
\(x=1\)
d,\(5^x\cdot5^{x+1}\cdot5^{x+2}< 1000000000000000000:2^{18}\)
\(5^{x+x+1+x+2}< 10^{18}:2^{18}\)
\(5^{3x+3}< 5^{18}\)
\(3x+3< 18\)
\(3\left(x+1\right)< 18\)
\(x+1< 6\)
\(x< 5\)
\(e,2^x\cdot\left(2^2\right)^2=\left(2^3\right)^2\)
\(2^x\cdot2^4=2^6\)
\(2^{x+4}=2^6\)
\(x+4=6\)
\(x=2\)
\(f,\left(x^5\right)^{10}=x\)
\(x^{50}=x\)
\(x^{50}-x=0\)
\(x\left(x^{49}-1\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
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a. 2^x.4=128
2^x=128:4=32
2^x=2^5
x=5
b. x^15=x
x=1
c. 2^x.(2^2)^2=(2^3)^2
2^x.2^4=2^6
2^(x+4)=2^6
x+4=6
x=6-4=2
d. (x^5)^10=x
x^50=x
x=1
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a) 2x-15=17
2x = 17+15 = 32
x = 32 : 2 =16
b) \(2^x.4=128\)
\(2^x=128:4=32\)
Mà \(32=2^5\)
Vậy x = 5
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a) \(2^x.4=128\)
\(2^x=128:4\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
b) \(x^{15}=x\)
\(\Rightarrow x\in\left\{0;1\right\}\)
c) \(2^x.\left(2^2\right)^2=\left(2^3\right)^2\)
\(2^x.2^4=2^6\)
\(2^x=2^6:2^4\)
\(2^x=2^2\)
\(\Rightarrow x=2\)
d) \(\left(x^5\right)^{10}=x\)
\(x^{50}=x\)
\(\Rightarrow x\in\left\{0;1\right\}\)
Ta có :
a) 2x . 4 = 128
=> 2x . 22 = 27
=> 2x = 27 : 22 = 25
=> x = 5
a, 2x.4=128
2x=128:4
2x=32
2x=25
x=5
Vay x=5
b, x15=x
x=1
Vay x=1
c, 2x.(22)2=(23)2
2x.16=64
2x=64:16
2x=4
2x=22
x=2
Vay x=2