Trata-se da operação que encontra o expoente de uma potência quando se conhece sua base e seu resultado, uma espécie de inverso da potenciação.

O logaritmo de um número é definido como o expoente ao qual se deve elevar a base a para obter o número x, ou seja

[img width=630,height=280]https://static.todamateria.com.br/upload/lo/ga/logaritmodefinicao-0.jpg[/img]

A função logarítmica de base a é definida como f (x) = loga x, com areal, positivo e a ≠ 1. A função inversa da função logarítmica é a função exponencial [br][br]

[math]f\left(x\right)=log_7x[/math][br][br][math]\text{g (x) = }log_{\frac{1}{3}}x+2[/math][br][br][math]h(x)=log_{10}x=logx[/math]

1°Propriedade[br][br]Quando não aparece a base de um logaritmo consideramos que seu valor é igual a 10[br][br]

Logaritmo de um produto:[br][br]Em qualquer base, o logaritmo do produto de dois ou mais números positivos é igual à soma dos logaritmos de cada um desses números.[br][br][img width=202,height=19]data:image/png;base64,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[/img][br][br]Exemplo Considerando[math]log_2=0,3[/math] e [math]log_3=0,48[/math], determine o valor do[math]log60[/math].[br][br]Solução[br][br]Podemos escrever o número 60 como um produto de 2.3.10. Neste caso, podemos aplicar a propriedade [br]para esse produto:[br][math]log60=log(2.3.10)[/math][br][br]Aplicando a propriedade do logaritmo de um produto:[br][math]log60=log2+log3+log10[/math][br][br]As bases são iguais a 10 e o [math]log10[/math] = 1. Substituindo esses valores, temos:[br][math]log60[/math] = 0,3 + 0,48 + 1 = 1,78

Logaritmo de um quociente:[br]Em qualquer base, o logaritmo do quociente de dois números reais e positivos é igual à diferença entre os logaritmos desses números.[br][br][img width=179,height=35]data:image/png;base64,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[/img][br][br]Exemplo:[br][br]Considerando [math]log5=0,70[/math], determine o valor do [math]log0,5[/math].[br][br]Solução[br][br]Podemos escrever 0,5 como sendo 5 dividido por 10, neste caso, podemos aplicar a propriedade do logaritmo de um quociente.[br][br][img width=180,height=110]data:image/png;base64,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[/img]

Logaritmo de uma potência:[br][br]Em qualquer base, o logaritmo de uma potência de base real e positiva é igual ao produto do expoente pelo logaritmo da base da potência.[br][br][img width=124,height=19]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAHwAAAATCAYAAABSvGkxAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAABGJhU0UAAAAOJ5y/mQAAAmpJREFUeNrtWEFEREEYflZWkuiUJJEOSbokSbolSZJIOnRIdOiURJKkQ6RDh65JkkTWSod0SdIpkqwkkU5JuiRJEq9v+OKZZrb33m67s9v8fHbnmTfz/98/8883z3Fy04qANeAdcIEKx1pe2xYwBEQsFeHNNcCHNx9JFLs5YTlMiUMjnK3xmcgBYMMmPCUOjXC2F9j20W8EmLUJT4lDrbPdwBXwwd8ezbstwBlF1C0wHCL4OWAGmAceONYRUK7w6RCI/jF5xcAScM/440BpnnCodLaZE9ezXc92i9Svls/b2K6mqArq7A7HmSKx4hxa1qzYGPDJOaJJyP8NyZJ9DiwAJZxjkskLknCTOfwxeEwRoGjvSs/WgdE0lLcboE56Jsh+yUJpXARW0lDSjeZQHvxFsXuiipefNKUuiLMRqkvZCrOQ8EiSmIIm3GgOXZ+TffjsF8TZZp41qnPtMAUBFaakNwEnaRJtRnMYdnXq7n1BnB1kWZNtIguKvIsCzcngDs8Kh6rzp0ch+2UyDhQipFTjbBWwzxUugu7j8xWqUpmYhE5h/qE1AtdpSrjRHLqK0nbrEQENbLcqdoQogWVcpZ3AKa8EcgB3QD/7jXmEUZyCo93zNS2uCCBTdkGFXkDRM66I2/VRwo3mUHeHvOZquuTq1H0M+b6vnjJQuWxN8l7rvTMO8f8jd8INr1sXSebKhFUCx/QlwficEAn/NxyWKYRCjCvZ2+50rOUkh3tckQ7Piz1FKRGX/w7PFeKV5dJaDnLYz9L3/VlwXNFHlJdN/hdl6NnmOP85FIJimqJlVfN1yVqeclhAWMsgh19zrwUZlRl4ugAAAM50RVh0TWF0aE1MADxtYXRoIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8xOTk4L01hdGgvTWF0aE1MIj48bXN1Yj48bWk+bG9nPC9taT48bWk+YTwvbWk+PC9tc3ViPjxtc3VwPjxtaT5iPC9taT48bWk+YzwvbWk+PC9tc3VwPjxtbz49PC9tbz48bWk+YzwvbWk+PG1vPi48L21vPjxtc3ViPjxtaT5sb2c8L21pPjxtaT5hPC9taT48L21zdWI+PG1pPmI8L21pPjwvbWF0aD7YNGC+AAAAAElFTkSuQmCC[/img][br][br]Podemos aplicar essa propriedade no logaritmo de uma raiz, pois, podemos escrever uma raiz na forma de expoente fracionário. Assim:[img width=217,height=43]data:image/png;base64,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[/img]Exemplo: Considerando [math]log3=0,48[/math], determine o valor do [math]log81[/math].[br][br]Solução:[br][br]Podemos escrever o número 81 como sendo 34. Neste caso, vamos aplicar a propriedade do logaritmo de uma potência, ou seja:[br][br][math]log81=log3^4[/math][br][math]log81=4.log3[/math][br][math]log81=4.0,48[/math][br][math]log81=1,92[/math]

Mudança de base:[br][br]Para aplicar as propriedades anteriores é necessário que todos os logaritmos da expressão estejam na mesma base. Do caso contrário, será necessário transformar todos para uma mesma base.[br][br]A mudança de base também é muito útil quando precisamos usar a calculadora para encontrar o valor de um logaritmo que está em uma base diferente de 10 e de e (base neperiana).[br][br]A mudança de base é feita aplicando-se a seguinte relação:[br][br][img width=111,height=44]data:image/png;base64,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[/img][br][br]Uma aplicação importante dessa propriedade é que o log ab é igual ao inverso do log ba, ou seja:[br][img width=112,height=40]data:image/png;base64,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[/img][br]Exemplo:[br][br]Escreva o [math]log_37[/math] na base 10.[br][br]Solução:[br][br]Vamos aplicar a relação para mudar o logaritmo para a base 10:[br][br][img width=107,height=38]data:image/png;base64,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[/img]

Para a construção do gráfico da função logarítmica devemos estar atentos a duas situações:[br][br][i][b] a > 1[br][br] 0 < a < 1[/b][/i]

O gráfico está totalmente à direita do eixo y, pois ela é definida para x > 0.[br][br]Intersecta o eixo das abscissas no ponto (1,0), então a raiz da função é x = 1.[br][br]Note que y assume todos as soluções reais, por isso dizemos que a Im(imagem) = R.[br][br][br]Através dos estudos das funções logarítmicas, chegamos à conclusão de que ela é uma função inversa da exponencial. Observe o gráfico comparativo a seguir:[br][img width=537,height=289]https://brasilescola.uol.com.br/upload/e/Untitled-3(47).jpg[/img][br][br]Podemos notar que (x,y) está no gráfico da função logarítmica se o seu inverso (y,x) está na função exponencial de mesma base.

1) [math]v(4)=log_2(4+4)=log_28=log_22^3[/math][br][math]v(4)=3.log_22[/math][br][math]v\left(4\right)=3\times1[/math][br][math]v(4)=3000,00[/math][br][br][math]v(t)=log_2(t+4)[/math][br][math]6=log_2(t+4)[/math][br][math]2^6=t+4[/math][br][math]t=64-4[/math][br][math]t=60[/math]

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