Question

Calcule o valor de x:
$\mathsf{log_x\,\dfrac{27\sqrt5}{25}=\dfrac{3}{2}}$​

Answer 1

Resposta:

x = 9/5

Explicação passo a passo:

Propriedade: $\displaystyle log_{a} b = c \Leftrightarrow a^c=b$

$\displaystyle log_{x} \frac{27\sqrt{5} }{25} = \frac{3}{2} \Leftrightarrow x^{\frac{3}{2} }=\frac{27\sqrt{5} }{25} = > \sqrt[2]{x^3} =\frac{3^3.5^{\frac{1}{2} }}{5^2}= > x^3 = (\frac{3^3.5^{\frac{1}{2} }}{5^2})^2\\\\\\x^3 = \frac{(3^3)^2.5^{(\frac{1}{2})^2 }}{(5^{2})^2}= \frac{(3^{3\times2}).5^{(\frac{1}{2}\times2) }}{5^{2\times2}}=\frac{3^6.5^1}{5^4}=\frac{3^6}{5^{(4-1)}} =\frac{3^6}{5^3} \\\\x = (\frac{3^6}{5^3})^{(\frac{1}{3} )}=\frac{3^{6\div3}}{5^{3\div3}} = \frac{3^2}{5^1} =\frac{9}{5}$

$Propriedades:(a^{m})^{n}=a^{m\times n}\\\\\sqrt[n]{x^m} =x^{\frac{m}{n} }\\\\a^{m}a^{n}=a^{m+n}\\\\\frac{a^{m}}{a^{n}}=a^{m-n} \\\\a^{0}=1\\\\a^{1}=a$