Question

resolva as equações exponenciais: me ajudem por favoooor ,urgenteeee!!!!!!​

Answer 1

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$\tt a)~\sf2^x\cdot4^{x+1}\cdot8^{x+2}=16^{x+3}\\\sf2^x\cdot(2^2)^{x+1}\cdot(2^3)^{x+2}=(2^4)^{x+3}\\\sf2^x\cdot2^{2x+2}\cdot2^{3x+6}=2^{4x+12}\\\sf2^{x+2x+2+3x+6}=2^{4x+12}\\\sf 2^{6x+8}=2^{4x+12}\\\sf 6x+8=4x+12\\\sf 6x-4x=12-8\\\sf 2x=4\\\sf x=\dfrac{4}{2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=2\checkmark}}}}$

$\tt b)~\sf 4^x-5\cdot2^x+4=0\\\sf (2^2)^x-5\cdot2^x+4=0\\\sf (2^x)^2-5\cdot2^x+4=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-5)^2-4\cdot1\cdot4\\\sf\Delta=25-16\\\sf\Delta=9\\\sf 2^x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf 2^x=\dfrac{-(-5)\pm\sqrt{9}}{2\cdot1}\\\sf2^x=\dfrac{5\pm3}{2}\begin{cases}\sf2^x=\dfrac{5+3}{2}=\dfrac{8}{2}=4\\\sf2^x=\dfrac{5-3}{2}=\dfrac{2}{2}=1\end{cases}\\\sf 2^x=4\\\sf 2^x=2^2\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=2}}}}\\\sf 2^x=1\\\sf 2^x=2^0\\\huge\boxed{\boxed{\boxed{\sf x=0}}}}$

$\tt c)~\sf 9^{x-2}=(\sqrt{27})^{3/4}\\\sf(3^2)^{x-2}=(\sqrt{3^3})^{3/4}\\\sf3^{2x-4}=(3^{\frac{3}{2}})^{\frac{3}{4}}\\\sf 3^{2x-4}=3^{\frac{9}{8}}\\\sf 2x-4=\dfrac{9}{8}\\\sf16x-32=9\\\sf16x=9+32\\\sf16x=41\\\huge\boxed{\boxed{\boxed{\sf x=\dfrac{41}{16}}}}$

$\tt d)~\sf2^{3x+1}=4^{x-2}\\\sf2^{3x+1}=(2^2)^{x-2}\\\sf3x+1=2x-4\\\sf3x-2x=-1-4\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=-5}}}}$

$\tt e)~\sf32^{x+2}=16^{x+1}\\\sf(2^5)^{x+2}=(2^4)^{x+1}\\\sf2^{5x+10}=2^{4x+4}\\\sf5x+10=4x+4\\\sf5x-4x=4-10\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=-6}}}}$