4 . 2x - 3 = 125
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\(\left(2x-1\right)^4=16=\left(\pm2\right)^4\\ =>\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left(2x+1\right)^3=125=5^3\\ =>2x+1=1\\ =>x=2\)
a) 2x . 4 = 128
<=> 2x = 32
<=> 2x = 25
<=> x = 5
b) x15 = x1
<=> x15 - x = 0
<=> x(x14 - 1) = 0
<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
c) (2x + 1)3 = 125
<=> (2x + 1)3 = 53
<=> 2x + 1 = 5
<=> 2x = 4
<=> x = 2
d) (x - 5)4 = (x - 5)6
<=> (x - 5)6 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)2 - 1] = 0
<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
Khi (x - 5)4 = 0 => x - 5 = 0 => x = 5
Khi (x - 5)2 - 1 = 0 <=> (x - 5)2 = 12 <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)
\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2.\\ b,\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}.\\ c,6.3^x-2.3^x=36\\ \Rightarrow3^x.\left(6-2\right)=36\\ \Rightarrow3^x.4=36\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2.\\ d,2^{x+1}-2^x=32\\ \Rightarrow2^x.\left(2-1\right)=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5.\)
4:
=>(2x+3,5)=7/12*3/14=21/168=1/8
=>2x=1/8-7/2=1/8-28/8=-27/8
=>x=-27/16
5: =>1/3:3x=-21/4
=>3x=-1/3:21/4=-1/3*4/21=-4/63
=>x=-4/189
6: =>2+7/9-3/4(x+1)=7/9
=>2-3/4(x+1)=0
=>3/4(x+1)=2
=>x+1=2:3/4=2*4/3=8/3
=>x=5/3
(2\(x\) - 1)3 = 125
(2\(x\) - 1)3 = 53
2\(x\) - 1 = 5
2\(x\) = 6
\(x\) = 3
(2\(x\) - 1)5 = \(x^5\)
2\(x\) - 1 = \(x\)
2\(x\) - \(x\) = 1
\(x\) = 1
2\(x^5\) + 2 = 4
2\(x^5\) = 4 - 2
2\(x^5\) = 2
\(x^5\) = 1
\(x\) = 1
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`(4 + 2x) * (9 - 3x) = 0`
TH1: `4 + 2x = 0 => 2x = -4 => x = -2`
TH2: `9 - 3x = 0 => 3x = 9 => x = 3`
Vậy, `x \in {-2; 3}`
____
`(3x - 19)^3 = 125`
`=> (3x - 19)^3 = 5^3`
`=> 3x - 19 = 5`
`=> 3x = 24`
`=> x = 8`
Vậy, `x = 8.`
\(4\cdot2^x-3=125\)
\(\Rightarrow4\cdot2^x=125+3\)
\(\Rightarrow4\cdot2^x=128\)
\(\Rightarrow2^x=128:4\)
\(\Rightarrow2^x=32\)
\(\Rightarrow x=5\)
\(\Leftrightarrow4.2^x=125+3=128\)
\(\Leftrightarrow2^x=128:4=32\)
\(\Rightarrow x=5\)