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\(3^x\cdot4=324\)
\(\Leftrightarrow3^x=\dfrac{324}{4}=81\)
\(\Leftrightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(3^x\cdot4=324\)
\(\Rightarrow3^x=324:4\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
`#3107.101107`
a)
\(5\left(x-1\right)^3=40\\\Rightarrow\left(x-1\right)^3=40\div5\\ \Rightarrow\left(x-1\right)^3=8\\ \Rightarrow\left(x-1\right)^3=2^3\\ \Rightarrow x-1=2\\ \Rightarrow x=2+1\\ \Rightarrow x=3\)
Vậy, `x = 3`
b)
\(3^{2x+1}+9^x=324?\\ \Rightarrow3^{2x}\cdot3+3^{2x}=324\\ \Rightarrow3^{2x}\cdot\left(3+1\right)=324\\ \Rightarrow3^{2x}\cdot4=324\\ \Rightarrow3^{2x}=81\\ \Rightarrow3^{2x}=3^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
c)
\(5^x-13=3\cdot2^2\\ \Rightarrow5^x-13=12\\ \Rightarrow5^x=12+13\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
Vậy, `x = 2`
d)
\(8^x+2^{3x+1}=192\\ \Rightarrow2^{3x}+2^{3x}\cdot2=192\\ \Rightarrow2^{3x}\left(1+2\right)=192\\ \Rightarrow2^{3x}\cdot3=192\\ \Rightarrow2^{3x}=64\\ \Rightarrow2^{3x}=2^6\\ \Rightarrow3x=6\\ \Rightarrow x=2\)
Vậy, `x = 2.`
+) \(24+\left(x-6\right)=135\)
\(\Rightarrow x-6=135-24=111\)
\(\Rightarrow x=111+6=117\)
+) \(2019-2\cdot\left(3x-2\right)=19\)
\(\Rightarrow2\cdot\left(3x-2\right)=2019-19=2000\)
\(\Rightarrow3x-2=2000:2=1000\)
\(\Rightarrow3x=1000+2=1002\)
\(\Rightarrow x=1002:3\)
\(\Rightarrow x=334\)
+) \(3^x+3^{x+1}=324\)
\(\Rightarrow3^x+3^x\cdot3=324\)
\(\Rightarrow3^x\cdot\left(1+3\right)=324\)
\(\Rightarrow3^x\cdot4=324\) \(\Rightarrow3^x=324:4=81\)
\(\Rightarrow3^x=3^4\) \(\Rightarrow x=4\)
3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹.(1 + 3 + 3²) = 39
3ˣ⁻¹.13 = 39
3ˣ⁻¹ = 39 : 13
3ˣ⁻¹ = 3
x - 1 = 1
x = 1 + 1
x = 2
\(3^{x-1}+3^x+3^{x+1}=39\)
\(=>3^x:3+3^x+3^x.3=39\)
\(=>3^x.\dfrac{1}{3}+3^x+3^x.3=39\)
\(=>3^x.\left(\dfrac{1}{3}+1+3\right)=39\)
\(=>3^x.\dfrac{13}{3}=39\)
\(=>3^x=39:\dfrac{13}{3}=39.\dfrac{3}{13}\)
\(=>3^x=9=3^2\)
\(=>x=2\)
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
Ta có:
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}\)
\(=x.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)\)
\(=x.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)\)
\(=x.\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(=x.\left(\frac{7}{14}-\frac{1}{14}\right)=x.\frac{3}{7}=\frac{1}{21}\)
\(\Rightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}.\frac{7}{3}=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
\(3^x+3^{x+1}=324\)
<=>\(3^x+3^x.3=324\)
<=>\(3^x.\left(1+3\right)=324\)
<=>\(3^x.4=324\)
<=>\(3^x=324:4\)
<=>\(3^x=81\)
<=>\(3^x=3^4\)
<=>\(x=4\)