What is 3ln(A)-1/2 [ln(B)+2ln(C^2)] ?

The solution to 3ln(A)-1/2 [ln(B)+2ln(C^2)] is 3ln(A)-1/2 ln(BC^4)

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simplify 3ln(A)-1/2 [ln(B)+2ln(C^2)]

simplify

3 ln (A) - \frac{1}{2} [ln (B) + 2 ln (C^{2})]

3 ln (A) - \frac{1}{2} ln (B C^{4})

Solution steps

3 ln (A) - \frac{1}{2} [ln (B) + 2 ln (C^{2})]

Apply log rule:

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

2 ln (C^{2}) = ln ((C^{2})^{2})

= 3 ln (A) - \frac{1}{2} [ln (B) + ln ((C^{2})^{2})]

Apply log rule:

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

ln (B) + ln ((C^{2})^{2}) = ln (B (C^{2})^{2})

= 3 ln (A) - \frac{1}{2} [ln (B (C^{2})^{2})]

(C^{2})^{2} = C^{4}

= 3 ln (A) - \frac{1}{2} [ln (B C^{4})]

Remove parentheses:

(a) = a

= 3 ln (A) - \frac{1}{2} ln (B C^{4})

- lo g_{10} (1.2 \cdot 1 0^{- 2})

ln (x) + ln (sin (x))

3 ln (10) - lo g_{10} (5) - lo g_{10} (2)

\frac{1}{3} ln (2 + 3)

- 2 ln (\frac{3}{4}) - ln (6)

What is 3ln(A)-1/2 [ln(B)+2ln(C^2)] ?
The solution to 3ln(A)-1/2 [ln(B)+2ln(C^2)] is 3ln(A)-1/2 ln(BC^4)