Tìm x biết \(\frac{3^2.3^8}{27^3}=3^x\)
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\(\frac{3^2.3^8}{27^3}=3^x\)
\(\Leftrightarrow\frac{3^{2+8}}{\left(3^3\right)^3}=3^x\)
\(\Leftrightarrow\frac{3^{10}}{3^9}=3^x\)
\(\Leftrightarrow3=3^x\)
\(\Leftrightarrow x=1\)
Ta có :\(\frac{3^2.3^8}{27^3}=3^x\)
\(\Rightarrow\frac{3^{10}}{\left(3^3\right)^3}=3^x\)
\(\Rightarrow\frac{3^{10}}{3^9}=3^x\)
\(\Rightarrow3^1=3^x\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
\(\frac{3^2.3^8}{37^3}=3^x\)
\(\Rightarrow\frac{3^{10}}{\left(3^3\right)^3}=3^x\Rightarrow\frac{3^{10}}{3^9}=3^x\Rightarrow3^1=3^x\Rightarrow x=1\)
a) \(\Leftrightarrow2.\left(\frac{2.3^x}{3}+3^x.3^2\right)=2.3^6\left(2+3^3\right)\)
\(\Leftrightarrow2.\left(\frac{2.3^x+3.3^x.3^2}{3}\right)=2.3^6.29\)
\(\Leftrightarrow2.\left[\frac{3^x.\left(2+3.3^2\right)}{3}\right]=2.3^6.19\)
\(\Leftrightarrow2.3^{x-1}.29=2.3^6.29\Leftrightarrow3^{x-1}.29=\frac{2.3^6.29}{2}=3^6.29\Leftrightarrow3^{x-1}=\frac{3^6.29}{29}=3^6\)
\(\Leftrightarrow3^{x-1}=3^6\Leftrightarrow x-1=6\Leftrightarrow x=6+1=7\)
vậy x=7 . Chọn mình nha
mấy bài sao tương tự nếu ko biết thì nhắn tin mình chỉ típ nha
a)
\(\text{( 25 – 2x )³ : 5 – 3^2 = 4^2}\)
\(\text{( 25 – 2x )³ : 5 – 9 = 16}\)
\(\text{( 25 – 2x )³ : 5 = 16 + 9}\)
\(\text{( 25 – 2x )³ : 5 = 25}\)
\(\text{( 25 – 2x )³ = 25 . 5}\)
\(\text{( 25 – 2x )³ = 125}\)
\(\text{( 25 – 2x )³ = 5³}\)
\(\text{25 – 2x = 5}\)
\(\text{2x = 25 – 5}\)
\(\text{2x = 20}\)
\(\text{x = 10}\)
\(\text{________________________________________}\)
b)
\(\text{2.3^x = 10.3^12 + 8.27^4}\)
\(\text{2.3^x = 10.3^12 + 8.(3^3)^4}\)
\(\text{2.3^x = 3^12 . (10+8)}\)
\(\text{2.3^x = 3^12 . 18}\)
\(\text{3^x = 3^12 . 18:2}\)
\(\text{3^x = 3^12 . 9}\)
\(\text{3^x = 3^12 . 3^2}\)
\(\text{3^x = 3^14}\)
\(\text{=> x=14}\)
\(\frac{3^2.3^8}{\left(3^3\right)^3}=3^x\frac{3^{10}}{3^9}=3^x3^{10-9}=3^x3^x=3x=1\)
\(\Rightarrow\frac{3^2.3^8}{3.3^8}=3^x\Rightarrow3=3^x\Rightarrow x=1\)