By the application of algebraic procedures and logarithm properties, the roots of the logarithmic equation [tex]\log \sqrt{x} + \log \sqrt{x-1} = 1[/tex] are x = 2 and x = -1.
In this question we need to solve a logarithmic equation by means of algebraic procedures and logarithm properties:
[tex]\log \sqrt{x} + \log \sqrt{x-1} = 1[/tex]
[tex]\log \sqrt{x\cdot (x-1)} = 1[/tex]
[tex]\log \sqrt{x^{2}-x} = 1[/tex]
[tex]\frac{1}{2}\cdot \log (x^{2}-x) = 1[/tex]
[tex]\log (x^{2}-x) = 2[/tex]
[tex]x^{2}-x = 2[/tex]
[tex]x \cdot (x-1) = 2[/tex]
[tex]x - 1 = \frac{2}{x}[/tex]
[tex]x - \frac{2}{x} = 1[/tex]
[tex]\frac{x^{2}-2}{x} = 1[/tex]
[tex]x^{2} - 2 = x[/tex]
[tex]x^{2}-x-2 = 0[/tex]
[tex](x-2)\cdot (x + 1) = 0[/tex]
[tex]x = 2 \,\lor \,x= -1[/tex]
By the application of algebraic procedures and logarithm properties, the roots of the logarithmic equation [tex]\log \sqrt{x} + \log \sqrt{x-1} = 1[/tex] are x = 2 and x = -1.
To learn more on logarithms: https://brainly.com/question/20785664
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