What is ln(|11.9-1|^2|2*11.9+1|^{1/2}) ?

The solution to ln(|11.9-1|^2|2*11.9+1|^{1/2}) is 2ln(10.9)+(ln(24.8))/2

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simplify ln(|11.9-1|^2|2*11.9+1|^{1/2})

simplify

ln (∣ 11.9 - 1 ∣^{2} ∣ 2 \cdot 11.9 + 1 ∣^{\frac{1}{2}})

2 ln (10.9) + \frac{ln ( 24.8 )}{2}

Decimal Notation

6.38294 \dots

Solution steps

ln (∣ 11.9 - 1 ∣^{2} ∣ 2 \cdot 11.9 + 1 ∣^{\frac{1}{2}})

Apply log rule:

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

ln (∣ 11.9 - 1 ∣^{2} ∣ 2 \cdot 11.9 + 1 ∣^{\frac{1}{2}}) = ln (∣ 11.9 - 1 ∣^{2}) + ln (∣ 2 \cdot 11.9 + 1 ∣^{\frac{1}{2}})

= ln (∣ 11.9 - 1 ∣^{2}) + ln (∣ 2 \cdot 11.9 + 1 ∣^{\frac{1}{2}})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (∣ 11.9 - 1 ∣^{2}) = 2 ln ∣ 11.9 - 1 ∣

= 2 ln ∣ 11.9 - 1 ∣ + ln (∣ 2 \cdot 11.9 + 1 ∣^{\frac{1}{2}})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (∣ 2 \cdot 11.9 + 1 ∣^{\frac{1}{2}}) = \frac{1}{2} ln ∣ 2 \cdot 11.9 + 1 ∣

= 2 ln ∣ 11.9 - 1 ∣ + \frac{1}{2} ln ∣ 2 \cdot 11.9 + 1 ∣

2 ln ∣ 11.9 - 1 ∣ = 2 ln (10.9)

\frac{1}{2} ln ∣ 2 \cdot 11.9 + 1 ∣ = \frac{ln ( 24.8 )}{2}

= 2 ln (10.9) + \frac{ln ( 24.8 )}{2}

7 lo g_{10} (y) + 2 lo g_{10} (5 y + 3)

0.7 ln (\frac{14 m}{14 x + 3 y})

- lo g_{10} (1.9 \cdot 1 0^{- 9})

ln (5.1 \cdot 1 0^{- 4}) + 0.94

3.74 + lo g_{10} (\frac{2.475}{0.025})

What is ln(|11.9-1|^2|2*11.9+1|^{1/2}) ?
The solution to ln(|11.9-1|^2|2*11.9+1|^{1/2}) is 2ln(10.9)+(ln(24.8))/2