What is simplify-1/2 ln|16-24+10|+1/2 ln|10| ?

The solution to simplify-1/2 ln|16-24+10|+1/2 ln|10| is 1/2 ln(5)

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simplify-1/2 ln|16-24+10|+1/2 ln|10|

simplify

- \frac{1}{2} ln ∣ 16 - 24 + 10 ∣ + \frac{1}{2} ln ∣ 10 ∣

\frac{1}{2} ln (5)

Decimal Notation

0.80471 \dots

Solution steps

- \frac{1}{2} ln ∣ 16 - 24 + 10 ∣ + \frac{1}{2} ln ∣ 10 ∣

Apply log rule:

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

\frac{1}{2} ln ∣ 16 + 10 - 24 ∣ = ln (∣ 16 + 10 - 24 ∣^{\frac{1}{2}})

= - ln (∣ 16 + 10 - 24 ∣^{\frac{1}{2}}) + \frac{1}{2} ln ∣ 10 ∣

Apply log rule:

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

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

= - ln (∣ 16 + 10 - 24 ∣^{\frac{1}{2}}) + ln (∣ 10 ∣^{\frac{1}{2}})

Apply log rule:

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

ln (∣ 10 ∣^{\frac{1}{2}}) - ln (∣ 16 - 24 + 10 ∣^{\frac{1}{2}}) = ln (\frac{∣ 10 ∣ ^{\frac{1}{2}}}{∣ 16 + 10 - 24 ∣ ^{\frac{1}{2}}})

= ln (\frac{∣ 10 ∣ ^{\frac{1}{2}}}{∣ 16 - 24 + 10 ∣ ^{\frac{1}{2}}})

\frac{∣ 10 ∣ ^{\frac{1}{2}}}{∣ 16 - 24 + 10 ∣ ^{\frac{1}{2}}} = 5

= ln (5)

Rewrite as

= ln (5^{\frac{1}{2}})

Apply log rule

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

assuming

x \geq 0

= \frac{1}{2} ln (5)

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What is simplify-1/2 ln|16-24+10|+1/2 ln|10| ?
The solution to simplify-1/2 ln|16-24+10|+1/2 ln|10| is 1/2 ln(5)