What is ln(e^x)+3ln(x)+4ln(x+1) ?

The solution to ln(e^x)+3ln(x)+4ln(x+1) is ln(e^xx^3(x+1)^4)

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simplify ln(e^x)+3ln(x)+4ln(x+1)

simplify

ln (e^{x}) + 3 ln (x) + 4 ln (x + 1)

ln (e^{x} x^{3} (x + 1)^{4})

Solution steps

ln (e^{x}) + 3 ln (x) + 4 ln (x + 1)

Apply log rule:

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

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

= ln (e^{x}) + ln (x^{3}) + 4 ln (x + 1)

Apply log rule:

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

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

= ln (e^{x} x^{3}) + 4 ln (x + 1)

Apply log rule:

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

4 ln (x + 1) = ln ((x + 1)^{4})

= ln (e^{x} x^{3}) + ln ((x + 1)^{4})

Apply log rule:

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

ln (e^{x} x^{3}) + ln ((x + 1)^{4}) = ln (e^{x} x^{3} (x + 1)^{4})

= ln (e^{x} x^{3} (x + 1)^{4})

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

5 ln (θ) + 5 ln (1 - θ)

ln (x^{3}) + 2 \cdot ln (x^{2})

ln (\frac{2 - 3 x}{3 + 2 x}) - 0

ln (10) - ln (70)

What is ln(e^x)+3ln(x)+4ln(x+1) ?
The solution to ln(e^x)+3ln(x)+4ln(x+1) is ln(e^xx^3(x+1)^4)