What is ln(3)(-1-1/2+1/8+3/4)+ln(2)(1-1/4-3/4) ?

The solution to ln(3)(-1-1/2+1/8+3/4)+ln(2)(1-1/4-3/4) is ln((3^{3/8})/3)

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simplify ln(3)(-1-1/2+1/8+3/4)+ln(2)(1-1/4-3/4)

simplify

ln (3) (- 1 - \frac{1}{2} + \frac{1}{8} + \frac{3}{4}) + ln (2) (1 - \frac{1}{4} - \frac{3}{4})

ln (\frac{3 ^{\frac{3}{8}}}{3})

Decimal Notation

- 0.68663 \dots

Solution steps

ln (3) (- 1 - \frac{1}{2} + \frac{1}{8} + \frac{3}{4}) + ln (2) (1 - \frac{1}{4} - \frac{3}{4})

Apply log rule:

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

ln (3) (\frac{1}{8} + \frac{3}{4} - 1 - \frac{1}{2}) = ln (3^{(\frac{1}{8} + \frac{3}{4} - 1 - \frac{1}{2})})

= ln (3^{(\frac{1}{8} + \frac{3}{4} - 1 - \frac{1}{2})}) + ln (2) (1 - \frac{1}{4} - \frac{3}{4})

Apply log rule:

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

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

= ln (3^{(\frac{1}{8} + \frac{3}{4} - 1 - \frac{1}{2})}) + ln (2^{(1 - \frac{1}{4} - \frac{3}{4})})

Apply log rule:

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

ln (3^{(- 1 - \frac{1}{2} + \frac{1}{8} + \frac{3}{4})}) + ln (2^{(1 - \frac{1}{4} - \frac{3}{4})}) = ln (3^{\frac{1}{8} + \frac{3}{4} - 1 - \frac{1}{2}} \cdot 2^{1 - \frac{1}{4} - \frac{3}{4}})

= ln (3^{\frac{1}{8} + \frac{3}{4} - 1 - \frac{1}{2}} \cdot 2^{1 - \frac{1}{4} - \frac{3}{4}})

3^{(- 1 - \frac{1}{2} + \frac{1}{8} + \frac{3}{4})} \cdot 2^{(1 - \frac{1}{4} - \frac{3}{4})} = \frac{3 ^{\frac{3}{8}}}{3}

= ln (\frac{3 ^{\frac{3}{8}}}{3})

\frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + C

4 ln (2 y) + 2 ln (y)

ln (2 ∣ ln (x) ∣)

ln (\frac{330}{293})

(5 \cdot 30) ln (\frac{( 2 \cdot 5 )}{30})

What is ln(3)(-1-1/2+1/8+3/4)+ln(2)(1-1/4-3/4) ?
The solution to ln(3)(-1-1/2+1/8+3/4)+ln(2)(1-1/4-3/4) is ln((3^{3/8})/3)