What is expand 1/2 ln|2(10)+1|+2ln|10-1| ?

The solution to expand 1/2 ln|2(10)+1|+2ln|10-1| is ln(81*21^{1/2})

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expand 1/2 ln|2(10)+1|+2ln|10-1|

expand

\frac{1}{2} ln ∣ 2 (10) + 1 ∣ + 2 ln ∣ 10 - 1 ∣

ln (81 \cdot 2 1^{\frac{1}{2}})

Decimal Notation

5.91671 \dots

Solution steps

\frac{1}{2} ln ∣ 2 (10) + 1 ∣ + 2 ln ∣ 10 - 1 ∣

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{2} ln ∣ 2 (10) + 1 ∣ = ln (∣ 2 (10) + 1 ∣^{\frac{1}{2}})

= ln (∣ 2 (10) + 1 ∣^{\frac{1}{2}}) + 2 ln ∣ 10 - 1 ∣

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln ∣ 10 - 1 ∣ = ln (∣ 10 - 1 ∣^{2})

= ln (∣ 2 (10) + 1 ∣^{\frac{1}{2}}) + ln (∣ 10 - 1 ∣^{2})

Apply log rule:

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

ln (∣ 2 (10) + 1 ∣^{\frac{1}{2}}) + ln (∣ 10 - 1 ∣^{2}) = ln (∣ 10 - 1 ∣^{2} ∣ 2 \cdot 10 + 1 ∣^{\frac{1}{2}})

= ln (∣ 10 - 1 ∣^{2} ∣ 2 \cdot 10 + 1 ∣^{\frac{1}{2}})

Expand

∣ 2 (10) + 1 ∣^{\frac{1}{2}} ∣ 10 - 1 ∣^{2} : 81 \cdot 2 1^{\frac{1}{2}}

= ln (81 \cdot 2 1^{\frac{1}{2}})

0.0591 \cdot ln (\frac{0.001}{0.1})

ln (5.5 \cdot 1 0^{- 15})

e^{\frac{l n ( 1 + 1 ^{4} ) - l n ( 17 )}{4}}

2 lo g_{0.5} (- 80 (x - 90)) + 10

10 lo g_{10} (\frac{2 \cdot 1 0 ^{- 7}}{1 0 ^{- 12}})

What is expand 1/2 ln|2(10)+1|+2ln|10-1| ?
The solution to expand 1/2 ln|2(10)+1|+2ln|10-1| is ln(81*21^{1/2})