It passes through (-1,4) and is perpendicular to the x axis

1 answer:

6 0

line x=4 is perpendicular to x-axis. So line, which is perpendicular to x=4 line, will be parallel to x-axis, therefore it is perpendicular to y-axis. Line, which is parallel to x-axis, has zero slope. Its equation, therefore, is y=0x+b or simply y=b. Since it shall pass through point (1,-4), its coordinates must satisfy the line equation. x does not matter, since there is no x in equation y=b, thus only y-coordinate matters. Plug in -4 into the equation, to obtain -4=b. So y=-4 is your answer. Answer: y=-4

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Expand using the properties and rules for logarithms

malfutka [58]

Consider expression $\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).$

1. Use property

$\log_a\dfrac{b}{c}=\log_ab-\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.$

2. Use property

$\log_abc=\log_ab+\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.$

3. Use property

$\log_ab^k=k\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.$

4. Use property

$\log_{a^k}b=\dfrac{1}{k}\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.$