What is log_{10}(-1/2 x)+1 ?

The solution to log_{10}(-1/2 x)+1 is log_{10}(x)+1-log_{10}(2)

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simplify log_{10}(-1/2 x)+1

simplify

lo g_{10} (- \frac{1}{2} x) + 1

lo g_{10} (x) + 1 - lo g_{10} (2)

Solution steps

lo g_{10} (- \frac{1}{2} x) + 1

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

lo g_{10} (- \frac{1}{2} x) = lo g_{10} (\frac{1}{2}) + lo g_{10} (x)

= lo g_{10} (\frac{1}{2}) + lo g_{10} (x) + 1

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

lo g_{10} (\frac{1}{2}) = lo g_{10} (1) - lo g_{10} (2)

= lo g_{10} (1) - lo g_{10} (2) + lo g_{10} (x) + 1

Apply log rule:

lo g_{a} (1) = 0

= 0 - lo g_{10} (2) + lo g_{10} (x) + 1

Group like terms

= lo g_{10} (x) - lo g_{10} (2) + 0 + 1

Add the numbers:

0 + 1 = 1

= lo g_{10} (x) + 1 - lo g_{10} (2)

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ln (\frac{e ^{x}}{x})

What is log_{10}(-1/2 x)+1 ?
The solution to log_{10}(-1/2 x)+1 is log_{10}(x)+1-log_{10}(2)