Qual é a função inversa da função g(x) = 4+5^x
Soluções para a tarefa
Respondido por
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Sabendo que
é uma função bijetora, concluímos que
possui inversa
e que por definição
![g\big(g^{-1}(x)\big)=x g\big(g^{-1}(x)\big)=x](https://tex.z-dn.net/?f=g%5Cbig%28g%5E%7B-1%7D%28x%29%5Cbig%29%3Dx)
Mas se
então,
![g\big(g^{-1}(x)\big)=4+5^{g^{-1}(x)}\\\\ x=4+5^{g^{-1}(x)}\\\\ 5^{g^{-1}(x)}=x-4 g\big(g^{-1}(x)\big)=4+5^{g^{-1}(x)}\\\\ x=4+5^{g^{-1}(x)}\\\\ 5^{g^{-1}(x)}=x-4](https://tex.z-dn.net/?f=g%5Cbig%28g%5E%7B-1%7D%28x%29%5Cbig%29%3D4%2B5%5E%7Bg%5E%7B-1%7D%28x%29%7D%5C%5C%5C%5C+x%3D4%2B5%5E%7Bg%5E%7B-1%7D%28x%29%7D%5C%5C%5C%5C+5%5E%7Bg%5E%7B-1%7D%28x%29%7D%3Dx-4)
Aplicando logaritmo de base 5 aos dois lados da igualdade acima, temos
![\mathrm{\ell og}_5\,\big(5^{g^{-1}(x)}\big)=\mathrm{\ell og}_5\,(x-4)\\\\ g^{-1}(x)\cdot \mathrm{\ell og}_5\,5=\mathrm{\ell og}_5\,(x-4)\\\\ g^{-1}(x)\cdot 1=\mathrm{\ell og}_5\,(x-4)\\\\\\ \therefore~~\boxed{\begin{array}{c} g^{-1}(x)=\mathrm{\ell og}_5\,(x-4) \end{array}}~~~~~~\text{com }x>4 \mathrm{\ell og}_5\,\big(5^{g^{-1}(x)}\big)=\mathrm{\ell og}_5\,(x-4)\\\\ g^{-1}(x)\cdot \mathrm{\ell og}_5\,5=\mathrm{\ell og}_5\,(x-4)\\\\ g^{-1}(x)\cdot 1=\mathrm{\ell og}_5\,(x-4)\\\\\\ \therefore~~\boxed{\begin{array}{c} g^{-1}(x)=\mathrm{\ell og}_5\,(x-4) \end{array}}~~~~~~\text{com }x>4](https://tex.z-dn.net/?f=%5Cmathrm%7B%5Cell+og%7D_5%5C%2C%5Cbig%285%5E%7Bg%5E%7B-1%7D%28x%29%7D%5Cbig%29%3D%5Cmathrm%7B%5Cell+og%7D_5%5C%2C%28x-4%29%5C%5C%5C%5C+g%5E%7B-1%7D%28x%29%5Ccdot+%5Cmathrm%7B%5Cell+og%7D_5%5C%2C5%3D%5Cmathrm%7B%5Cell+og%7D_5%5C%2C%28x-4%29%5C%5C%5C%5C+g%5E%7B-1%7D%28x%29%5Ccdot+1%3D%5Cmathrm%7B%5Cell+og%7D_5%5C%2C%28x-4%29%5C%5C%5C%5C%5C%5C+%5Ctherefore%7E%7E%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D+g%5E%7B-1%7D%28x%29%3D%5Cmathrm%7B%5Cell+og%7D_5%5C%2C%28x-4%29+%5Cend%7Barray%7D%7D%7E%7E%7E%7E%7E%7E%5Ctext%7Bcom+%7Dx%26gt%3B4)
Mas se
Aplicando logaritmo de base 5 aos dois lados da igualdade acima, temos
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