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64 . 4x = 168
<=> 43. 4x = 416
=> 3 + x = 16
<=> x = 13
Vậy x = 13
2x.162 = 1024
<=> 2x. 28 = 210
=> x + 8 = 10
<=> x = 2
Vậy x = 2
b: Ta có: \(2^x\cdot16^2=1024\)
\(\Leftrightarrow2^x\cdot2^8=2^{10}\)
\(\Leftrightarrow x+8=10\)
hay x=2
2x.162 = 1024
=> 2x.(24)2 = 210
=> 2x + 4.2 = 10
=> x + 8 = 10
=> x = 2
vậy_
64.4x = 168
=> 43.4x = (42)8
=> 43 + x = 416
=> 3 + x = 16
=> x = 13
vậy_
2x = 16
=> 2x = 24
=> x = 4
vậy_
\(f\)) \(32^{-x}.16^x=1024\)
\(\left(2\right)^{-5x}.2^{4x}=2^{10}\)
\(\Leftrightarrow2^{4x-5x}=2^{10}\)
\(\Leftrightarrow2^{-x}=2^{10}\)
\(\Leftrightarrow-x=10\)
\(\Leftrightarrow x=-10\)
\(g\)) \(3^{x-1}.5+3^{x-1}=162\)
\(3^{x-1}.\left(5+1\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(h\)) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=\left(1,-1\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=-1\\2x-1=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=0\\2x=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=0\\x=1\end{cases}}\)
\(i\)) \(5^x+5^{x+2}=650\)
\(5^x.\left(1+5^2\right)=650\)
\(5^x.26=650\)
\(5^x=650:26\)
\(5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)
32x . 16x = 1024
=> (25)x . (24)x = 210
=> 25x . 24x = 210
=> 25x + 4x = 210
=> 29x = 210
=> 9x = 10
=> x = 10 : 9
=> x = 10/9
TL:
\(32^x.16^x=1024\)
\(\left(2^5\right)^x.\left(2^4\right)^x=1024\)
\(2^{9x}=2^{10}\)
\(\Rightarrow9x=10\Leftrightarrow x=\frac{10}{9}\)
Vậy............