Question

Resolva em R as seguintes equações exponenciais:

Answer 1

$\large\boxed{\begin{array}{l}\sf Resolva\,em\,\mathbb{R}\,as\,seguintes\,equac_{\!\!,}\tilde oes\,exponenciais\!:\\\sf a)\,3^x=243\\\sf b)\,5^{x^2-8x+15}=1\\\sf c)\, 4^x=\dfrac{1}{32}\\\\\sf d)\,9^x=\sqrt[\sf3]{\sf81}\\\sf e)\, 9^{2x-1}=27^{5x+1}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf Soluc_{\!\!,}\tilde ao\!:}\\\rm a) 3^x=243\\\rm 3^x=3^5\\\rm x=5\\\rm S=\{5\}\\\rm b)\,5^{x^2-8x+15}=1\\\rm 5^{x^2-8x+15}=5^0\\\rm x^2-8x+15=0\\\rm \Delta=b^2-4ac\\\rm\Delta=8^2-4\cdot1\cdot15\\\rm\Delta=64-60\\\rm\Delta=4\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\rm x=\dfrac{-(-8)\pm\sqrt{4}}{2\cdot1}\\\\\rm x=\dfrac{8\pm2}{2}\begin{cases}\rm x_1=\dfrac{8+2}{2}=\dfrac{10}{2}=5\\\\\rm x_2=\dfrac{8-2}{2}=\dfrac{6}{2}=3\end{cases}\\\\\rm S=\{3,5\}\end{array}}$

$\large\boxed{\begin{array}{l}\rm c)\,4^x=\dfrac{1}{32}\\\\\rm (2^2)^{x}=2^{-5}\\\rm 2^{2x}=2^{-5}\\\rm 2x=-5\\\rm x=-\dfrac{5}{2}\\\\\rm S=\bigg\{-\dfrac{5}{2}\bigg\}\\\\\rm d)\, 9^x=\sqrt[\rm3]{\rm81}\\\rm (3^2)^x=\sqrt[\rm3]{\rm3^4}\\\rm 3^{2x}=3^{\frac{4}{3}}\\\rm 2x=\dfrac{4}{3}\\\\\rm 6x=4\\\rm x=\dfrac{4\div2}{6\div2}\\\\\rm x=\dfrac{2}{3}\\\\\rm S=\bigg\{\dfrac{2}{3}\bigg\}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm e)\, 9^{2x-1}=27^{5x+1}\\\rm (3^2)^{2x-1}=(3^3)^{5x+1}\\\rm 3^{4x-2}=3^{15x+3}\\\rm 4x-2=15x+3\\\rm 15x+3=4x-2\\\rm 15x-4x=-2-3\\\rm 11x=-5\\\rm x=-\dfrac{5}{11}\\\\\rm S=\bigg\{\!\!-\dfrac{5}{11}\bigg\}\end{array}}$