Question

Write each expression as a sum and/or difference of logarithms. Express powers as factors. $$ \ln \left[\frac{5 x^{2} \sqrt[3]{1-x}}{4(x+1)^{2}}\right] 0

Short answer

\[ \ln(5) + 2 \ln(x) + \frac{1}{3} \ln(1-x) - \ln(4) - 2 \ln(x+1) \]

Step-by-step solution

Apply the logarithm of a quotient

Use the logarithm property \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \) to split the logarithm into two separate logs.\[\ln\left(\frac{5 x^{2} \sqrt[3]{1-x}}{4(x+1)^{2}\right) = \ln\left(5 x^{2} \sqrt[3]{1-x}\right) - \ln\left(4(x+1)^{2}\right)\]

Apply the logarithm of a product

Use the logarithm property \( \ln(ab) = \ln(a) + \ln(b)\) to split both resulting logarithms into simpler components.\[\ln\left(5 x^{2} \sqrt[3]{1-x}\right) - \ln\left(4 (x+1)^{2} \right) = \ln(5) + \ln(x^{2}) + \ln(\sqrt[3]{1-x}) - \ln(4) - \ln((x+1)^2) \]

Apply the logarithm of a power

Use the logarithm property \( \ln(a^n)= n \ln(a) \) to simplify terms involving powers.\[ \ln(5) + \ln(x^{2}) + \ln(\sqrt[3]{1-x}) - \ln(4) - \ln((x+1)^2) = \ln(5) + 2 \ln(x) + \frac{1}{3} \ln(1-x) - \ln(4) - 2 \ln(x+1)\]

Key concepts

Logarithm of a Quotient

The logarithm of a quotient property is very useful for breaking down complex logarithmic expressions. This property states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator. Mathematically, it is represented as: \( \ln\left( \frac{a}{b} \right)=\ln(a)-\ln(b) \).
Applying this property helps to simplify expressions by separating them into smaller, more manageable parts.
For instance, in our exercise, we start with
\ \ln\left( \frac{5 x^{2} \sqrt[3]{1-x}}{4(x+1)^{2}} \right) = \ln\left( 5 x^{2} \sqrt[3]{1-x} \right)- \ln\left( 4(x+1)^{2} \right)
This first step simplifies a complex fraction into two simpler logarithmic expressions that can be further manipulated.

Logarithm of a Product

The logarithm of a product property allows us to split a single logarithm that contains a product into the sum of separate logarithms. This property is written as: \( \ln(ab) =\ln(a)+\ln(b) \).
It is highly useful for breaking down products into their individual components.
In our exercise, after splitting the quotient using the first property, we apply this property to both resulting components, yielding:
\( \ln(5 x^{2} \sqrt[3]{1-x} )- \ln\left( 4 ( x+1 )^{2} \right) = \ln(5)+\ln( x^{2})+ \ln(\sqrt[3]{1-x})-\ln(4)- \ln((x+1)^{2}) \)
By breaking the products into sums, we create several simpler logarithmic expressions that are much easier to handle.

Logarithm of a Power

The logarithm of a power property helps to remove exponents from inside a logarithm by turning them into multipliers. This property states: \( \ln( a^n )=n\ln(a) \).
This allows us to further simplify expressions containing powers.
In our exercise, this property is applied to terms with exponents within the logarithms. For instance:
\( \ln(5)+\ln( x^{2})+ \ln(\sqrt[3]{1-x})-\ln(4)- \ln((x+1)^{2}) \) becomes:
\( \ln(5)+2 \ln(x)+\frac{1}{3} \ln(1-x)-\ln(4)-2 \ln(x+1) \)
By applying this property, complex exponents are converted into simpler multiplicative factors, making the expression much easier to interpret and work with.