What is ln((e^{3x}(x+1))/(2x^{12)}) ?

The solution to ln((e^{3x}(x+1))/(2x^{12)}) is 3x+ln(x+1)-ln(2)-12ln(x)

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simplify ln((e^{3x}(x+1))/(2x^{12)})

simplify

ln (\frac{e ^{3 x} ( x + 1 )}{2 x ^{12}})

3 x + ln (x + 1) - ln (2) - 12 ln (x)

Solution steps

ln (\frac{e ^{3 x} ( x + 1 )}{2 x ^{12}})

Apply log rule:

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

ln (\frac{e ^{3 x} ( x + 1 )}{2 x ^{12}}) = ln (e^{3 x} (x + 1)) - ln (2 x^{12})

= ln (e^{3 x} (x + 1)) - ln (2 x^{12})

Apply log rule:

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

ln (e^{3 x} (x + 1)) = ln (e^{3 x}) + ln (x + 1), ln (2 x^{12}) = ln (2) + ln (x^{12})

= ln (e^{3 x}) + ln (x + 1) - (ln (x^{12}) + ln (2))

Apply log rule:

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

ln (e^{3 x}) = 3 x ln (e)

= 3 x ln (e) + ln (x + 1) - (ln (2) + ln (x^{12}))

Apply log rule:

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

ln (x^{12}) = 12 ln (x)

= 3 x ln (e) + ln (x + 1) - (ln (2) + 12 ln (x))

3 x ln (e) = 3 x

= 3 x + ln (x + 1) - (12 ln (x) + ln (2))

- (ln (2) + 12 ln (x)) : - ln (2) - 12 ln (x)

= 3 x + ln (x + 1) - ln (2) - 12 ln (x)

3 \cdot lo g_{10} (x - 3) - 4.5 lo g_{10} (x - 2)

\frac{ln ( \frac{30}{7} )}{3 ln ( 3 )}

(ln ∣ 2 ∣ - ln ∣ 1 ∣) - (5) - (ln ∣ 4 ∣ - ln ∣ 3 ∣)

lo g_{10} (z x^{6})

- \frac{1}{0.00012} ln (\frac{0.211}{1})

What is ln((e^{3x}(x+1))/(2x^{12)}) ?
The solution to ln((e^{3x}(x+1))/(2x^{12)}) is 3x+ln(x+1)-ln(2)-12ln(x)