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Resolvendo a equação:
![2^x+2^{x+1}=3^x\\\\\\ 2^x+2^x \times 2=3^x\ (\ Colocar\ o\ 2^x\ em\ evid\^encia\ )\\\\\\ 2^x(1+2)=3^x\\\\\\ 2^x \times 3=3^x\\\\\\ 2^x=\dfrac{3^x}{3}\\\\\\ 2=\sqrt[x]{\dfrac{3^x}{3}}\\\\\\ 2=\dfrac{\sqrt[x]{3^x}}{\sqrt[x]{3}}}\\\\\\ 2=\dfrac{\sqrt[\not x]{3^{\not x}}}{\sqrt[x]{3}}}\\\\\\ 2=\dfrac{3}{\sqrt[x]{3}}}\\\\\\ \sqrt[x]{3}=\dfrac{3}{2}\\\\\\ 3=\left(\dfrac{3}{2}\right)^x\\\\\\ \boxed{\boxed{x=\log_{\frac{3}{2}}3}}](https://tex.z-dn.net/?f=2%5Ex%2B2%5E%7Bx%2B1%7D%3D3%5Ex%5C%5C%5C%5C%5C%5C+2%5Ex%2B2%5Ex+%5Ctimes+2%3D3%5Ex%5C+%28%5C+Colocar%5C+o%5C+2%5Ex%5C+em%5C+evid%5C%5Eencia%5C+%29%5C%5C%5C%5C%5C%5C+2%5Ex%281%2B2%29%3D3%5Ex%5C%5C%5C%5C%5C%5C+2%5Ex+%5Ctimes+3%3D3%5Ex%5C%5C%5C%5C%5C%5C+2%5Ex%3D%5Cdfrac%7B3%5Ex%7D%7B3%7D%5C%5C%5C%5C%5C%5C+2%3D%5Csqrt%5Bx%5D%7B%5Cdfrac%7B3%5Ex%7D%7B3%7D%7D%5C%5C%5C%5C%5C%5C+2%3D%5Cdfrac%7B%5Csqrt%5Bx%5D%7B3%5Ex%7D%7D%7B%5Csqrt%5Bx%5D%7B3%7D%7D%7D%5C%5C%5C%5C%5C%5C+2%3D%5Cdfrac%7B%5Csqrt%5B%5Cnot+x%5D%7B3%5E%7B%5Cnot+x%7D%7D%7D%7B%5Csqrt%5Bx%5D%7B3%7D%7D%7D%5C%5C%5C%5C%5C%5C+2%3D%5Cdfrac%7B3%7D%7B%5Csqrt%5Bx%5D%7B3%7D%7D%7D%5C%5C%5C%5C%5C%5C+%5Csqrt%5Bx%5D%7B3%7D%3D%5Cdfrac%7B3%7D%7B2%7D%5C%5C%5C%5C%5C%5C+3%3D%5Cleft%28%5Cdfrac%7B3%7D%7B2%7D%5Cright%29%5Ex%5C%5C%5C%5C%5C%5C+%5Cboxed%7B%5Cboxed%7Bx%3D%5Clog_%7B%5Cfrac%7B3%7D%7B2%7D%7D3%7D%7D "2^x+2^{x+1}=3^x\\\\\\ 2^x+2^x \times 2=3^x\ (\ Colocar\ o\ 2^x\ em\ evid\^encia\ )\\\\\\ 2^x(1+2)=3^x\\\\\\ 2^x \times 3=3^x\\\\\\ 2^x=\dfrac{3^x}{3}\\\\\\ 2=\sqrt[x]{\dfrac{3^x}{3}}\\\\\\ 2=\dfrac{\sqrt[x]{3^x}}{\sqrt[x]{3}}}\\\\\\ 2=\dfrac{\sqrt[\not x]{3^{\not x}}}{\sqrt[x]{3}}}\\\\\\ 2=\dfrac{3}{\sqrt[x]{3}}}\\\\\\ \sqrt[x]{3}=\dfrac{3}{2}\\\\\\ 3=\left(\dfrac{3}{2}\right)^x\\\\\\ \boxed{\boxed{x=\log_{\frac{3}{2}}3}}")
Espero ter ajudado.
Bons estudos!
Espero ter ajudado.
Bons estudos!
MAranda:
Eu tinha feito de outra forma, mas mesmo assim sua forma parece mais elegante. Obrigado por compartilhar seu conhecimento :)
Que bom. Fico feliz. Por nada. :)
Obrigado :D
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