1) \(2^x-15=17\)
\(\Leftrightarrow2^x=17+15\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
2) \(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=32\cdot25+200\)
\(\Leftrightarrow\left(7x-11\right)^3=800+200\)
\(\Leftrightarrow\left(7x-11\right)^3=1000\)
\(\Leftrightarrow\left(7x-11\right)^3=10^3\)
\(\Leftrightarrow7x-11=10\)
\(\Leftrightarrow7x=10+11\)
\(\Leftrightarrow7x=21\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
3) \(3^x+25=26\cdot2^2+2\cdot3^0\)
\(\Leftrightarrow3^x+25=26\cdot4+2\cdot1\)
\(\Leftrightarrow3^x+25=106\)
\(\Leftrightarrow3^x=106-25\)
\(\Leftrightarrow3^x=81\)
\(\Leftrightarrow x=4\)
4) \(49\cdot7^x=2401\) (có sửa lại)
\(\Leftrightarrow7^x=49\)
\(\Leftrightarrow7^x=7^2\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
5) \(3^x=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
