2x-26=6
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`2^x = 16`
`=> 2^x = 2^4`
`=> x = 4`
Vậy, `x = 4.`
____
`2^x*16 = 1024`
`=> 2^x =`\(2^{10}\div2^4\)
`=> 2^x = 2^6`
`=> x = 6`
Vậy, `x = 6`
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`2^x - 26 = 6`
`=> 2^x = 6 + 26`
`=> 2^x = 32`
`=> 2^x = 2^5`
`=> x = 5`
Vậy, `x = 5`
`3^x*3 = 243`
`=> 3^x * 3 = 3^5`
`=> 3^x = 3^5 \div 3`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4.`
\(2^x-26=6\)
\(2^x=6+26\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
\(#WendyDang\)
( 26 - 12 )+( x + 5 ) = 14 - ( 2x - 6 )
14 + ( x + 5 ) = 14 - ( 2x - 6 )
=> x + 5 = 2x - 6
x = 2x - 6 - 5
x = 2x - 1
\(\Leftrightarrow x^2-6x+8=6\sqrt{2x+1}-18\left(Đk:x\ge-\dfrac{1}{2}\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=\dfrac{12\left(x-4\right)}{\sqrt{2x+1}+3}\left(\sqrt{2x+1}+3>0\right)\)
+) \(x=4\left(TM\right)\)
+) \(x\ne4\Rightarrow x-2=\dfrac{12}{\sqrt{2x+1}+3}\)
\(\Leftrightarrow x-4=\dfrac{12-2\left(\sqrt{2x+1}+3\right)}{\sqrt{2x+1}+3}\)
\(\Leftrightarrow x-4+\dfrac{2\left(x-4\right)}{\left(\sqrt{2x+1}+3\right)^2}=0\)
\(\Leftrightarrow1+\dfrac{2}{\left(\sqrt{2x+1}+3\right)^2}=0\left(x\ne4\right)\)
Vì \(\dfrac{2}{\left(\sqrt{2x+1}+3\right)^2}>0\forall x\) => VT>0
=> phương trình vô nghiệm
Vậy \(S=\left\{4\right\}\)
ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
\(\Leftrightarrow\left(x^2-8x+16\right)+\left(2x+1-6\sqrt{2x+1}+9\right)=0\)
\(\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{2x+1}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\\sqrt{2x+1}-3=0\end{matrix}\right.\) \(\Leftrightarrow x=4\)
2ˣ - 26 = 6
2ˣ = 6 + 26
2ˣ = 32
2ˣ = 2⁵
x = 5
\(2^x-26=6\Leftrightarrow2^x=6+26=32\)
\(\Rightarrow2^x=2^5\Leftrightarrow x=5\)