Bài 1: Tìm x.
a) \(2^3.2^x=64\)
b) \(7.7^x=343\)
c) \(7.7^{x+1}=343\)
d) \(2^x-15=17\)
e) \(x^{10}=1\)
g) \(x^{10}=x\)
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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 )
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) \(7.7^{x+1}=343\)
\(\Rightarrow7^{x+1}=343\div7\)
\(\Rightarrow7^{x+1}=49\)
\(\Rightarrow7^{x+1}=7^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=2-1\)
\(\Rightarrow x=1\)
a) \(\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\Rightarrow\frac{28}{27}=x:\frac{3}{7}\Rightarrow x=\frac{4}{9}\)
b) \(\left(x-\frac{4}{7}\right)^3=343\Rightarrow\left(x-\frac{4}{7}\right)^3=7^3\Rightarrow x-\frac{4}{7}=7\Rightarrow x=\frac{53}{7}\)
c) \(x^5=x^3\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)
e) \(\left(x-1\right)^4=16\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^4=2^4\\\left(x-1\right)^4=\left(-2\right)^4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
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
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) 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
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)