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a) \(2^3.2^x=64\Leftrightarrow8.2^x=64\Leftrightarrow2^x=\dfrac{64}{8}=8=2^3\Rightarrow x=3\)
vậy \(x=3\)
b) \(7.7^x=343\Leftrightarrow7^x=\dfrac{343}{7}=49=7^2\Rightarrow x=2\)
vậy \(x=2\)
c) \(7.7^{x+1}=343\Leftrightarrow7^{x+1}=\dfrac{343}{7}=49=7^2\Rightarrow x+1=2\Leftrightarrow x=1\)
vậy \(x=1\)
d) \(2^x-15=17\Leftrightarrow2^x=17+15=32=2^5\Rightarrow x=5\)
vậy \(x=5\)
e) \(x^{10}=1\Leftrightarrow x^{10}=1^{10}\Rightarrow x=1\)
vậy \(x=1\)
\(x^{10}=x\Leftrightarrow x^{10}-x=0\Leftrightarrow x\left(x^9-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy \(x=0;x=1\)
a) \(2^3.2^x=2^3.2^3\)
=> \(2^x=2^3\)
=> x=3
b) \(7.7^x=343\)
=> \(7.7^x=7.7^2\)
=> \(7^x=7^2\)
=> x=2
c) \(7.7^{x+1}=7.7^2\)
=> \(7^{x+1}=7^2\)
=> x+1=2
=> x=1
d) \(2^x-15=17\)
=> \(2^x=17+15\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> x=5
e) \(x^{10}=1\)
=> \(x^{10}=1^{10}\)
=> x=1
g) \(x^{10}=x\)
=> x = 0 hoặc 1
+) Với x=0 => \(0^{10}=0\)( t/m )
+) Với x=1 => \(1^{10}=1\)( t/m )
a) \(25\le5^x\le125\)
\(\Rightarrow5^2\le5^x\le5^3\)
\(\Rightarrow x\in\left\{2;3\right\}\)
b) \(7.7^{x+1}=343\)
\(\Rightarrow7^{x+1}=343\div7\)
\(\Rightarrow7^{x+1}=49\)
\(\Rightarrow7^{x+1}=7^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=2-1\)
\(\Rightarrow x=1\)
1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4
a)
7x - 2x = 617 : 615 + 44 : 11
=> 5x = 62 + 4
=> 5x = 36 + 4
=> 5x = 40
=> x = 40 : 5
=> x = 8
Vậy x = 8
b)
2x+1 . 22014 = 22015
=> 2x+1 = 22015 : 22014
=> 2x+1 = 21
=> x+1 = 1
=> x = 1-1
=> x = 0
c)
\(|\)x-2\(|\) = 0
=> x-2 = 0
=> x = 0+2
=> x = 2
e) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\cdot\left(2x-15\right)^2-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left(2x-15\right)^3=0\) hoặc \(\left(2x-15\right)^2-1=0\)
+)TH1: \(\left(2x-15\right)^3=0\)
\(\Rightarrow2x-15=0\)
\(\Rightarrow2x=15\)
\(\Rightarrow x=\frac{15}{2}\)
+)TH2: \(\left(2x-15\right)^2-1=0\)
\(\Rightarrow\left(2x-15\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=1\\2x-15=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=16\\2x=14\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
Vậy \(x=\frac{15}{2}\) hoặc \(x=8\) hoặc \(x=7\)
a) \(2^x-17=15\Rightarrow2^x=32\)
Mà \(2^5=32\Rightarrow x=5\)
Vậy x = 5
b)\(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
Vậy x = 3
c)\(x^{10}=1^x\Rightarrow x^{10}=1\)(số 1 có luỹ thừa là bao nhiêu thì vẫn là 1 thui)\(\Rightarrow x=1\)
Vậy x = 1
d) \(x^{10}=x\Rightarrow x^{10}-x=0\)
\(\Rightarrow x\left(x^9-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x^9-1=0\)
+)TH1: \(x=0\)
+)TH2: \(x^9-1=0\Rightarrow x^9=1\Rightarrow x=1\)
Vậy x = 0 hoặc x = 1
a) 2x - 15 = 17
2x = 25
b) ( 7x - 11 )3 = 25. 52 + 200
(7x - 11)3 = 32 . 25 + 200
(7x -11)3 = 1000
(7x-11)3 = 103
7x - 11 = 10
7x = 10+11
7x = 21
x = 21 : 7
x = 3
like nha
a, -5/6 -x = 7/12 + -1/3
⇔-10/12 - 12x/12 = 7/12 + -4/12
⇒-10 - 12x = 7 - 4
⇔-12x = 7 - 4 +10
⇔-12x = 13
⇔x = -13/12
b, x+13/-15 = 1/3
⇔-(x+13)/15 = 5/15
⇒ -x - 13 = 5
⇔-x = 5 +13
⇔-x = 18
⇔x = -18
c,-15/x-1 = -3/5
⇔-75/(x-1).5 = -3.(x-1)/5.(x-1)
⇒-75 = -3x + 3
⇔3x = 3 + 75
⇔3x = 78
⇔x = 26
d, (1/2).x + -2/5 = 1/5
⇔5x/10 + -4/10 = 1/10
⇒5x - 4 = 1
⇔5x = 1 + 4
⇔5x = 5
⇔x = 1
e, (-2/3).x + 1/5 = 1/10
⇔-20x/30 + 6/30 = 3/30
⇒-20x + 6 = 3
⇔-20x = 3 - 6
⇔-20x = -3
⇔x = 3/20
f, 4/5 - (1/2).x = 1/10
⇔8/10 - 5x/10 = 1/10
⇒8 - 5x = 1
⇔-5x = 1 - 8
⇔-5x = -7
⇔x=7/5
a) 2x=64:23=64:8=8=23=>x=3
b) 7x=343:7=49=72=> x=2
c) Tương tự như câu trên, với x+1 thì x=1
d) 2x=17+15=32=25=>x=5
e) => \(x\in\left(-1;1\right)\)
g) =>x=0;1
Ta có : x10 = x
=> x10 - x = 0
=> x(x9 - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)