Question

resolva a equação exponencial a seguir:
$2 {}^{x} = 32 \\ \\ 5 {}^{x} { }^{ - 2 } = 125 \\ \\ 3 {}^{x} {}^{ + 3} = 243 \\ \\ 7 {}^{x {}^{2} } {}^{ - 5x + 6 } = 1$
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Answer 1

$\large\boxed{\begin{array}{l}\sf2^x=32\\\begin{array}{c|c}\sf32&\sf2\\\sf16&\sf2\\\sf8&\sf2\\\sf4&\sf2\\\sf2&\sf2\\\sf1\end{array}\\\sf 32=2^5\\\sf 2^x=2^5\\\sf x=2\\\sf S=\{2\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf5^{x-2}=125\\\begin{array}{c|c}\sf125&\sf5\\\sf25&\sf5\\\sf5&\sf5\\\sf1\end{array}\\\sf 125=5^3\\\sf5^{x-2}=5^3\\\sf x-2=3\\\sf x=2+3\\\sf x=5\\\sf S=\{5\}\end{array}}$

$\boxed{\begin{array}{l}\sf 3^{x+3}=243\\\begin{array}{c|c}\sf243&\sf3\\\sf81&\sf3\\\sf27&\sf3\\\sf9&\sf3\\\sf3&\sf3\\\sf1\end{array}\\\sf 243=3^5\\\sf 3^{x+3}=3^5\\\sf x+3=5\\\sf x=5-3\\\sf x=2\\\sf S=\{2\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf7^{x^2-5x+6}=1\\\sf 7^{x^2-5x+6}=7^0\\\sf x^2-5x+6=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-5)^2-4\cdot1\cdot6\\\sf\Delta=25-24\\\sf\Delta=1\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-5)\pm\sqrt{1}}{2\cdot1}\\\\\sf x=\dfrac{5\pm1}{2}\begin{cases}\sf x_1=\dfrac{5+1}{2}=\dfrac{6}{2}=3\\\\\sf x_2=\dfrac{5-1}{2}=\dfrac{4}{2}=2\end{cases}\\\sf S=\{2,3\}\end{array}}$