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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
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
a) \(3^x=27=3^3\)
=> x= 3
b) \(5^x=125=5^3\)
=> x = 3
c) \(2^x.2^3=2^{64}\)
\(2^{x+3}=2^6\)
=> x+ 3 = 6
=> x = 6 - 3
=> x = 3
d) \(7.7^{x+2}=343\)
\(7^{x+3}=7^3\)
=> x + 3 = 3
=> x = 0
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.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
a) 23.2n = 26
<=> 23+n = 26
<=> 3+n=6
<=> n = 3
b) 7.7n+1 = 343
<=> 71+n+1 = 73
<=> 72+n = 73
<=> 2+n=3
<=> n = 1
a)23*2n=26
\(\Rightarrow2^{3+n}=2^6\)
\(\Rightarrow3+n=6\Rightarrow n=3\)
b)\(7\cdot7^{n+1}=343\)
\(\Rightarrow7^{n+2}=7^3\)
\(\Rightarrow n+2=3\Rightarrow n=1\)
a) 3x = 81 = 34
=> x = 4
b) 5x = 125 = 53
=> x = 3
c) 7.7x+1 = 343
=> 7x . 49 = 343
=> 7x = 7
=> x = 1
d) 2.3x = 162
=> 3x = 81 =34
=> x = 4
e) 2x - 15 = 17
2x = 32 = 25
=> x = 5
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 )