What is (1/2)^x-1/4 ln(1/x)+arcsin(x)-5/4 ?

The solution to (1/2)^x-1/4 ln(1/x)+arcsin(x)-5/4 is (1/2)^x+1/4 ln(x)+arcsin(x)-5/4

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simplify (1/2)^x-1/4 ln(1/x)+arcsin(x)-5/4

simplify

(\frac{1}{2})^{x} - \frac{1}{4} ln (\frac{1}{x}) + arcsin (x) - \frac{5}{4}

(\frac{1}{2})^{x} + \frac{1}{4} ln (x) + arcsin (x) - \frac{5}{4}

Solution steps

(\frac{1}{2})^{x} - \frac{1}{4} ln (\frac{1}{x}) + arcsin (x) - \frac{5}{4}

Apply log rule:

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

ln (\frac{1}{x}) = ln (1) - ln (x)

= (\frac{1}{2})^{x} - \frac{1}{4} (- ln (x) + ln (1)) + arcsin (x) - \frac{5}{4}

\frac{1}{4} (ln (1) - ln (x)) = - \frac{1}{4} ln (x)

= (\frac{1}{2})^{x} - (- \frac{1}{4} ln (x)) + arcsin (x) - \frac{5}{4}

Apply rule

- (- a) = a

= (\frac{1}{2})^{x} + \frac{1}{4} ln (x) + arcsin (x) - \frac{5}{4}

ln (13) - ln (21)

ln (y) - ln (x) + 1

- lo g_{10} (2.1 \cdot 1 0^{- 10})

ln (\frac{10}{16})

2.49 - \frac{0.0592}{1} \cdot lo g_{10} (\frac{0.35}{0.25})

What is (1/2)^x-1/4 ln(1/x)+arcsin(x)-5/4 ?
The solution to (1/2)^x-1/4 ln(1/x)+arcsin(x)-5/4 is (1/2)^x+1/4 ln(x)+arcsin(x)-5/4