What is 2ln(x)+5ln(2)-ln(8) ?

The solution to 2ln(x)+5ln(2)-ln(8) is ln(4x^2)

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simplify 2ln(x)+5ln(2)-ln(8)

simplify

2 ln (x) + 5 ln (2) - ln (8)

ln (4 x^{2})

Solution steps

2 ln (x) + 5 ln (2) - ln (8)

Apply log rule:

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

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

= ln (x^{2}) + 5 ln (2) - ln (8)

Apply log rule:

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

5 ln (2) = ln (2^{5})

= ln (x^{2}) + ln (2^{5}) - ln (8)

Apply log rule:

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

ln (x^{2}) + ln (2^{5}) = ln (2^{5} x^{2})

= ln (2^{5} x^{2}) - ln (8)

2^{5} = 32

= ln (32 x^{2}) - ln (8)

Apply log rule:

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

ln (32 x^{2}) - ln (8) = ln (\frac{32 x ^{2}}{8})

= ln (\frac{32 x ^{2}}{8})

Divide the numbers:

\frac{32}{8} = 4

= ln (4 x^{2})

7 ln (x) + 2 ln (3) - 5 ln (y)

\frac{1}{3} lo g_{10} (x) - lo g_{10} (5)

3 ln ∣ x ∣ - 2 ln ∣ x + 1 ∣

ln (a^{4} b^{7})

\frac{1}{2} lo g_{7} (x) + \frac{1}{3} lo g_{7} (y)