$$ \operatorname { Ln } ( x ^ { 3 } ( 2 x - 5 ) ! ) $$

Question

UpStudy AI · Accepted Answer

To simplify the expression $$ \operatorname{Ln}(x^3 (2x - 5)!) $$, we can use the properties of logarithms.

1. **Apply the logarithm property**: The logarithm of a product can be expressed as the sum of the logarithms:
   $$
   \operatorname{Ln}(a \cdot b) = \operatorname{Ln}(a) + \operatorname{Ln}(b)
   $$
   Therefore, we can rewrite the expression as:
   $$
   \operatorname{Ln}(x^3) + \operatorname{Ln}((2x - 5)!)
   $$

2. **Simplify $$ \operatorname{Ln}(x^3) $$**: Using the power rule of logarithms:
   $$
   \operatorname{Ln}(x^3) = 3 \cdot \operatorname{Ln}(x)
   $$

3. **Combine the results**: The expression now becomes:
   $$
   3 \cdot \operatorname{Ln}(x) + \operatorname{Ln}((2x - 5)!)
   $$

Thus, the simplified form of the expression $$ \operatorname{Ln}(x^3 (2x - 5)!) $$ is:
$$
3 \cdot \operatorname{Ln}(x) + \operatorname{Ln}((2x - 5)!)
$$
The simplified form of $$ \operatorname{Ln}(x^3 (2x - 5)!) $$ is $$ 3 \cdot \operatorname{Ln}(x) + \operatorname{Ln}((2x - 5)!) $$.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Calculus Questions