Como calcular essa expressão usando logaritmos?
1,1^n=40
Considere log 11=1,04 1 log 2=0,3
Soluções para a tarefa
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Dadas as propriedades:
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Adote:
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= = = = =
Dada a expressão:
Aplicando a propriedade logarítmica em ambos os termos:
![\mathsf{\log \left(1,1^n\right)=\log \left(40\right)}\\\\\\\mathsf{n\cdot \log \left(1,1\right)=\log \left(40\right)}\\\\\\\mathsf{n\cdot \log \left(\dfrac{11}{10}\right)=\log \left(4\cdot 10\right)}\\\\\\\mathsf{n\cdot \log \left(\dfrac{11}{10}\right)=\log \left(2^2\cdot 10\right)}\\\\\\\mathsf{n\cdot \left[\log \left(11\right)-\log \left(10\right)\right]=\log \left(2^2\right)+\log \left(10\right)}\\\\\\\mathsf{n\cdot \left[\log \left(11\right)-1\right]=\left[2\cdot \:\log \left(2\right)\right]+1}\\\\\\\mathsf{n\cdot \left(1,04-1\right)=\left(2\cdot \:0,3\right)+1}\\\\\\\mathsf{n\cdot 0,04=0,6+1}\\\\\\\mathsf{n\cdot 0,04=1,6}\\\\\\\mathsf{n=\dfrac{1,6}{0,04}}\\\\\\\mathsf{n=\dfrac{160}{4}}\\\\\\\boxed{\mathsf{n=40}}\: \: \checkmark](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Clog+%5Cleft%281%2C1%5En%5Cright%29%3D%5Clog+%5Cleft%2840%5Cright%29%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%5Ccdot+%5Clog+%5Cleft%281%2C1%5Cright%29%3D%5Clog+%5Cleft%2840%5Cright%29%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%5Ccdot+%5Clog+%5Cleft%28%5Cdfrac%7B11%7D%7B10%7D%5Cright%29%3D%5Clog+%5Cleft%284%5Ccdot+10%5Cright%29%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%5Ccdot+%5Clog+%5Cleft%28%5Cdfrac%7B11%7D%7B10%7D%5Cright%29%3D%5Clog+%5Cleft%282%5E2%5Ccdot+10%5Cright%29%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%5Ccdot+%5Cleft%5B%5Clog+%5Cleft%2811%5Cright%29-%5Clog+%5Cleft%2810%5Cright%29%5Cright%5D%3D%5Clog+%5Cleft%282%5E2%5Cright%29%2B%5Clog+%5Cleft%2810%5Cright%29%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%5Ccdot+%5Cleft%5B%5Clog+%5Cleft%2811%5Cright%29-1%5Cright%5D%3D%5Cleft%5B2%5Ccdot+%5C%3A%5Clog+%5Cleft%282%5Cright%29%5Cright%5D%2B1%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%5Ccdot+%5Cleft%281%2C04-1%5Cright%29%3D%5Cleft%282%5Ccdot+%5C%3A0%2C3%5Cright%29%2B1%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%5Ccdot+0%2C04%3D0%2C6%2B1%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%5Ccdot+0%2C04%3D1%2C6%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%3D%5Cdfrac%7B1%2C6%7D%7B0%2C04%7D%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bn%3D%5Cdfrac%7B160%7D%7B4%7D%7D%5C%5C%5C%5C%5C%5C%5Cboxed%7B%5Cmathsf%7Bn%3D40%7D%7D%5C%3A+%5C%3A+%5Ccheckmark "\mathsf{\log \left(1,1^n\right)=\log \left(40\right)}\\\\\\\mathsf{n\cdot \log \left(1,1\right)=\log \left(40\right)}\\\\\\\mathsf{n\cdot \log \left(\dfrac{11}{10}\right)=\log \left(4\cdot 10\right)}\\\\\\\mathsf{n\cdot \log \left(\dfrac{11}{10}\right)=\log \left(2^2\cdot 10\right)}\\\\\\\mathsf{n\cdot \left[\log \left(11\right)-\log \left(10\right)\right]=\log \left(2^2\right)+\log \left(10\right)}\\\\\\\mathsf{n\cdot \left[\log \left(11\right)-1\right]=\left[2\cdot \:\log \left(2\right)\right]+1}\\\\\\\mathsf{n\cdot \left(1,04-1\right)=\left(2\cdot \:0,3\right)+1}\\\\\\\mathsf{n\cdot 0,04=0,6+1}\\\\\\\mathsf{n\cdot 0,04=1,6}\\\\\\\mathsf{n=\dfrac{1,6}{0,04}}\\\\\\\mathsf{n=\dfrac{160}{4}}\\\\\\\boxed{\mathsf{n=40}}\: \: \checkmark")
๏
๏
๏
๏
= = = = =
Adote:
๏
๏
= = = = =
Dada a expressão:
Aplicando a propriedade logarítmica em ambos os termos:
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