log7 (4) +log7 (2x^3-1 ) = log7 (339)
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Essa questão terá como resultado a fração de Sete Meios, veja a comprovação:
Neste caso temos a soma e igualdade de logaritmos, devemos então inicialmente utilizar a seguinte propriedade: log (a) + log (b) = log (a • b), veja:
![\mathsf{log_{7}^~{4}}+\mathsf{log_{7}^~{(2x^{3}-1)}}}} = \mathsf{log_{7}^~{339}}}\\\\\\\\\\ \mathsf{log_{7}^~{4}}~\cdot~(2x^{3}-1)}}} = \mathsf{log_{7}^~{339}}}~~~\textsf{Aplicando~o~ln~em~ambos~os~lados:}}\\\\\\\\\\ \mathsf{4~\cdot~(2x^{3}-1) = 339}}}\\\\\\\\\\ \mathsf{2x^{3}-1= \dfrac{339}{4}}}\\\\\\\\\\ \mathsf{2x^{3} = \dfrac{339+4}{4}}}\\\\\\\\\ \mathsf{x^{3} = \dfrac{339+4}{4}}~\cdot~{\dfrac{1}{2}}}}\\\\\\\\\\ \mathsf{x = \sqrt[3]{\dfrac{343}{8}}}}\\\\\\\\\ \mathsf{x = \dfrac{\sqrt[3]{343}}{\sqrt[3]{8}}}}}}\\\\\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{x=\dfrac{7}{2}}}}}}}}}}}}}}}}}}}}}}}}](https://tex.z-dn.net/?f=+%5Cmathsf%7Blog_%7B7%7D%5E%7E%7B4%7D%7D%2B%5Cmathsf%7Blog_%7B7%7D%5E%7E%7B%282x%5E%7B3%7D-1%29%7D%7D%7D%7D+%3D+%5Cmathsf%7Blog_%7B7%7D%5E%7E%7B339%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Blog_%7B7%7D%5E%7E%7B4%7D%7D%7E%5Ccdot%7E%282x%5E%7B3%7D-1%29%7D%7D%7D+%3D+%5Cmathsf%7Blog_%7B7%7D%5E%7E%7B339%7D%7D%7D%7E%7E%7E%5Ctextsf%7BAplicando%7Eo%7Eln%7Eem%7Eambos%7Eos%7Elados%3A%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B4%7E%5Ccdot%7E%282x%5E%7B3%7D-1%29+%3D+339%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B2x%5E%7B3%7D-1%3D+%5Cdfrac%7B339%7D%7B4%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B2x%5E%7B3%7D+%3D+%5Cdfrac%7B339%2B4%7D%7B4%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bx%5E%7B3%7D+%3D+%5Cdfrac%7B339%2B4%7D%7B4%7D%7D%7E%5Ccdot%7E%7B%5Cdfrac%7B1%7D%7B2%7D%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bx+%3D+%5Csqrt%5B3%5D%7B%5Cdfrac%7B343%7D%7B8%7D%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bx+%3D+%5Cdfrac%7B%5Csqrt%5B3%5D%7B343%7D%7D%7B%5Csqrt%5B3%5D%7B8%7D%7D%7D%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Clarge%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cmathsf%7Bx%3D%5Cdfrac%7B7%7D%7B2%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D+ "\mathsf{log_{7}^~{4}}+\mathsf{log_{7}^~{(2x^{3}-1)}}}} = \mathsf{log_{7}^~{339}}}\\\\\\\\\\ \mathsf{log_{7}^~{4}}~\cdot~(2x^{3}-1)}}} = \mathsf{log_{7}^~{339}}}~~~\textsf{Aplicando~o~ln~em~ambos~os~lados:}}\\\\\\\\\\ \mathsf{4~\cdot~(2x^{3}-1) = 339}}}\\\\\\\\\\ \mathsf{2x^{3}-1= \dfrac{339}{4}}}\\\\\\\\\\ \mathsf{2x^{3} = \dfrac{339+4}{4}}}\\\\\\\\\ \mathsf{x^{3} = \dfrac{339+4}{4}}~\cdot~{\dfrac{1}{2}}}}\\\\\\\\\\ \mathsf{x = \sqrt[3]{\dfrac{343}{8}}}}\\\\\\\\\ \mathsf{x = \dfrac{\sqrt[3]{343}}{\sqrt[3]{8}}}}}}\\\\\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{x=\dfrac{7}{2}}}}}}}}}}}}}}}}}}}}}}}}")
Espero que te ajude '-'
Neste caso temos a soma e igualdade de logaritmos, devemos então inicialmente utilizar a seguinte propriedade: log (a) + log (b) = log (a • b), veja:
Espero que te ajude '-'
jpcg26:
Ok,obrigado
Respondido por
0
Boa tarde Jpcg
log7(4) + log7(2x³ - 1) = log7(339)
log7(8x³ - 4) = log7(339)
cancelando os logs fica
8x³ - 4 = 339
8x³ = 339 + 4 = 343
x³ = 343/8
x³ = 7³/2³
x = 7/2
log7(4) + log7(2x³ - 1) = log7(339)
log7(8x³ - 4) = log7(339)
cancelando os logs fica
8x³ - 4 = 339
8x³ = 339 + 4 = 343
x³ = 343/8
x³ = 7³/2³
x = 7/2
Ok, obrigado
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