a) Ta có: \(15.3^x-1=1215\)
\(3^{x-1}=81=3^4\)
\(x-1=4\)
\(x=5\)
Vậy \(x=5\)
b) Ta có: \(\left(2x-1\right).3=27\)
\(2x-1=9\)
\(2x=10\)
\(x=5\)
Vậy \(x=5\)
a)
\(15\cdot3^{x-1}=1215\)
\(3^{x-1}=1215:15\)
\(3^{x-1}=81\)
\(3^{x-1}=3^4\)
\(x-1=4\)
\(x=4+1\)
\(x=5\)
b)
\(\left(2x-1\right)\cdot3=27\)
\(2x-1=27:3\)
\(2x-1=9\)
\(2x=9+1\)
\(2x=10\)
\(x=10:2\)
\(x=5\)
c)
\(\left(x+1\right)^2=64\)
\(x+1=\pm\sqrt{64}=\pm8\)
\(\orbr{\begin{cases}x+1=8\\x+1=-8\end{cases}}\)
\(\orbr{\begin{cases}x=8-1\\x=-8-1\end{cases}}\)
\(\orbr{\begin{cases}x=7\\x=-9\end{cases}}\)
Bài làm :
\(a\text{)}...\Leftrightarrow3^{x-1}=\frac{1215}{15}=81\Leftrightarrow3^{x-1}=3^4\Leftrightarrow x-1=4\Leftrightarrow x=5\)
\(b\text{)}...\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Leftrightarrow x=5\)
\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=8^2\\\left(x+1\right)^2=\left(-8\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+1=8\\x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=7\\x=-9\end{cases}}\)