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3 years ago

Which function is equivalent to f(x) = lnx?

Mathematics

2 answers:

11Alexandr11 [23.1K]3 years ago

7 0

Answer:

$f(x)=log_ex$

Step-by-step explanation:

By definition, we can write ln instead of log. WHEN??

Whenever the base of the logarithm is the number "e".

Hence, when we have:

$Log_e$

We can write it in shortcut as:

$Log_e=ln$

Hence, ln x can also be written as $Log_ex$

Fourth answer choice is right.

luda_lava [24]3 years ago

4 0

Answer:

$f(x) = log_e(x)$

Step-by-step explanation:

$f(x) = ln x$

For logarithmic function ln(x) the base of ln is 'e'

$f(x) = log_3(x)$

The base of log is 3 . so it is not equivalent to $f(x) = lnx$

$f(x) = log_{10}(x)$

The base of log is 10 . so it is not equivalent to $f(x) = lnx$

$f(x) = log_b(x)$

The base of log is b . so it is not equivalent to $f(x) = lnx$

$f(x) = log_e(x)$

The base of log is e . so it is equivalent to $f(x) = lnx$

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3 years ago

Custom Office makes a line of executive desks. It is estimated that the total cost for making x units of their Senior Executive

Ivan

Answer:

(a) The average cost function is $\bar{C}(x)=95+\frac{230000}{x}$

(b) The marginal average cost function is $\bar{C}'(x)=-\frac{230000}{x^2}$

(c) The average cost approaches to 95 if the production level is very high.

Step-by-step explanation:

(a) Suppose $C(x)$ is a total cost function. Then the average cost function, denoted by $\bar{C}(x)$ , is

$\frac{C(x)}{x}$

We know that the total cost for making x units of their Senior Executive model is given by the function

$C(x) = 95x + 230000$

The average cost function is

$\bar{C}(x)=\frac{C(x)}{x}=\frac{95x + 230000}{x} \\\bar{C}(x)=95+\frac{230000}{x}$

(b) The derivative $\bar{C}'(x)$ of the average cost function, called the marginal average cost function, measures the rate of change of the average cost function with respect to the number of units produced.

The marginal average cost function is

$\bar{C}'(x)=\frac{d}{dx}\left(95+\frac{230000}{x}\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g\\\\\frac{d}{dx}\left(95\right)+\frac{d}{dx}\left(\frac{230000}{x}\right)\\\\\bar{C}'(x)=-\frac{230000}{x^2}$

(c) The average cost approaches to 95 if the production level is very high.

$\lim_{x \to \infty} (\bar{C}(x))=\lim_{x \to \infty} (95+\frac{230000}{x})\\\\\lim _{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim _{x\to a}f\left(x\right)\pm \lim _{x\to a}g\left(x\right)\\\\=\lim _{x\to \infty \:}\left(95\right)+\lim _{x\to \infty \:}\left(\frac{230000}{x}\right)\\\\\lim _{x\to a}c=c\\\lim _{x\to \infty \:}\left(95\right)=95\\\\\mathrm{Apply\:Infinity\:Property:}\:\lim _{x\to \infty }\left(\frac{c}{x^a}\right)=0\\\lim_{x \to \infty} (\frac{230000}{x} )=0$