Log√5 X =(9/4)
Calcule o valor de X
Atenção:√5 é a base
Soluções para a tarefa
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Definição de logaritmo:
Propriedades utilizadas:
![\boxed{\boxed{\sqrt[n]{a^{m}}=a^{(m/n)}}}\\\\\\\boxed{\boxed{log_{b}(a)=\dfrac{log_{c}(a)}{log_{c}(b)}~~(mudanca~de~base~de~b~para~c)}}\\\\\\\boxed{\boxed{log_{(b^{n})}(a)=\dfrac{1}{n}\cdot log_{b}(a)~~~~(vem~da~propriedade~acima)}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Csqrt%5Bn%5D%7Ba%5E%7Bm%7D%7D%3Da%5E%7B%28m%2Fn%29%7D%7D%7D%5C%5C%5C%5C%5C%5C%5Cboxed%7B%5Cboxed%7Blog_%7Bb%7D%28a%29%3D%5Cdfrac%7Blog_%7Bc%7D%28a%29%7D%7Blog_%7Bc%7D%28b%29%7D%7E%7E%28mudanca%7Ede%7Ebase%7Ede%7Eb%7Epara%7Ec%29%7D%7D%5C%5C%5C%5C%5C%5C%5Cboxed%7B%5Cboxed%7Blog_%7B%28b%5E%7Bn%7D%29%7D%28a%29%3D%5Cdfrac%7B1%7D%7Bn%7D%5Ccdot+log_%7Bb%7D%28a%29%7E%7E%7E%7E%28vem%7Eda%7Epropriedade%7Eacima%29%7D%7D "\boxed{\boxed{\sqrt[n]{a^{m}}=a^{(m/n)}}}\\\\\\\boxed{\boxed{log_{b}(a)=\dfrac{log_{c}(a)}{log_{c}(b)}~~(mudanca~de~base~de~b~para~c)}}\\\\\\\boxed{\boxed{log_{(b^{n})}(a)=\dfrac{1}{n}\cdot log_{b}(a)~~~~(vem~da~propriedade~acima)}}")
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![log_{(\sqrt{5})}(x)=\dfrac{9}{4}\\\\\\log_{(\sqrt[2]{5^{1}})}(x)=\dfrac{9}{4}\\\\\\log_{[5^{(1/2)}]}(x)=\dfrac{9}{4}\\\\\\\dfrac{1}{(\frac{1}{2})}\cdot log_{5}(x)=\dfrac{9}{4}\\\\\\2\cdot log_{5}(x)=\dfrac{9}{4}\\\\\\log_{5}(x)=\dfrac{9}{4}\cdot\dfrac{1}{2}\\\\\\\boxed{\boxed{log_{5}(x)=\dfrac{9}{8}}}](https://tex.z-dn.net/?f=log_%7B%28%5Csqrt%7B5%7D%29%7D%28x%29%3D%5Cdfrac%7B9%7D%7B4%7D%5C%5C%5C%5C%5C%5Clog_%7B%28%5Csqrt%5B2%5D%7B5%5E%7B1%7D%7D%29%7D%28x%29%3D%5Cdfrac%7B9%7D%7B4%7D%5C%5C%5C%5C%5C%5Clog_%7B%5B5%5E%7B%281%2F2%29%7D%5D%7D%28x%29%3D%5Cdfrac%7B9%7D%7B4%7D%5C%5C%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B%28%5Cfrac%7B1%7D%7B2%7D%29%7D%5Ccdot+log_%7B5%7D%28x%29%3D%5Cdfrac%7B9%7D%7B4%7D%5C%5C%5C%5C%5C%5C2%5Ccdot+log_%7B5%7D%28x%29%3D%5Cdfrac%7B9%7D%7B4%7D%5C%5C%5C%5C%5C%5Clog_%7B5%7D%28x%29%3D%5Cdfrac%7B9%7D%7B4%7D%5Ccdot%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5Cboxed%7B%5Cboxed%7Blog_%7B5%7D%28x%29%3D%5Cdfrac%7B9%7D%7B8%7D%7D%7D "log_{(\sqrt{5})}(x)=\dfrac{9}{4}\\\\\\log_{(\sqrt[2]{5^{1}})}(x)=\dfrac{9}{4}\\\\\\log_{[5^{(1/2)}]}(x)=\dfrac{9}{4}\\\\\\\dfrac{1}{(\frac{1}{2})}\cdot log_{5}(x)=\dfrac{9}{4}\\\\\\2\cdot log_{5}(x)=\dfrac{9}{4}\\\\\\log_{5}(x)=\dfrac{9}{4}\cdot\dfrac{1}{2}\\\\\\\boxed{\boxed{log_{5}(x)=\dfrac{9}{8}}}")
Aplicando a definição de logaritmos:
Então:
![x=5^{(9/8)}\\\\x=5^{(8/8)+(1/8)}\\\\x=5^{(8/8)}\cdot5^{(1/8)}\\\\x=5^{1}\cdot\sqrt[8]{5^{1}}\\\\\boxed{\boxed{x=5\sqrt[8]{5}}}](https://tex.z-dn.net/?f=x%3D5%5E%7B%289%2F8%29%7D%5C%5C%5C%5Cx%3D5%5E%7B%288%2F8%29%2B%281%2F8%29%7D%5C%5C%5C%5Cx%3D5%5E%7B%288%2F8%29%7D%5Ccdot5%5E%7B%281%2F8%29%7D%5C%5C%5C%5Cx%3D5%5E%7B1%7D%5Ccdot%5Csqrt%5B8%5D%7B5%5E%7B1%7D%7D%5C%5C%5C%5C%5Cboxed%7B%5Cboxed%7Bx%3D5%5Csqrt%5B8%5D%7B5%7D%7D%7D "x=5^{(9/8)}\\\\x=5^{(8/8)+(1/8)}\\\\x=5^{(8/8)}\cdot5^{(1/8)}\\\\x=5^{1}\cdot\sqrt[8]{5^{1}}\\\\\boxed{\boxed{x=5\sqrt[8]{5}}}")
Propriedades utilizadas:
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Aplicando a definição de logaritmos:
Então:
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