Find the derivative of the following function: $y(x)=(\mathrm{ln}(2x){)}^{5x},\phantom{\rule{1em}{0ex}}x>\frac{1}{2}$

Pellagra3d
2022-10-24
Answered

Find the derivative of the following function: $y(x)=(\mathrm{ln}(2x){)}^{5x},\phantom{\rule{1em}{0ex}}x>\frac{1}{2}$

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cdtortosadn

Answered 2022-10-25
Author has **19** answers

Whenever you have a function of the form $y=f(x{)}^{g(x)}$, it is usually a good idea to use logarithmic differentiation:

$$\mathrm{log}y=g(x)\mathrm{log}f(x),$$

hence

$$\frac{d}{dx}[\mathrm{log}y]=\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[g(x)\mathrm{log}f(x)],$$

and

$$\frac{dy}{dx}=f(x{)}^{g(x)}(\frac{g(x)}{f(x)}\frac{df}{dx}+\frac{dg}{dx}\mathrm{log}f(x)).$$

So, in your case, $f(x)=\mathrm{log}2x$ and $g(x)=5x$, so we have

$$\frac{df}{dx}=\frac{1}{2x}\cdot 2=\frac{1}{x},$$

and

$$\frac{dg}{dx}=5,$$

therefore after minor simplifications,

$$\frac{dy}{dx}=(\mathrm{log}2x{)}^{5x}(\frac{5}{\mathrm{log}2x}+5\mathrm{log}(\mathrm{log}2x)).$$

The only additional step that was done was to put the fractions over a common denominator of $\mathrm{log}2x$ and factoring:

$$\frac{dy}{dx}=5(\mathrm{log}2x{)}^{5x-1}(1+(\mathrm{log}2x)\mathrm{log}(\mathrm{log}2x)).$$

$$\mathrm{log}y=g(x)\mathrm{log}f(x),$$

hence

$$\frac{d}{dx}[\mathrm{log}y]=\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[g(x)\mathrm{log}f(x)],$$

and

$$\frac{dy}{dx}=f(x{)}^{g(x)}(\frac{g(x)}{f(x)}\frac{df}{dx}+\frac{dg}{dx}\mathrm{log}f(x)).$$

So, in your case, $f(x)=\mathrm{log}2x$ and $g(x)=5x$, so we have

$$\frac{df}{dx}=\frac{1}{2x}\cdot 2=\frac{1}{x},$$

and

$$\frac{dg}{dx}=5,$$

therefore after minor simplifications,

$$\frac{dy}{dx}=(\mathrm{log}2x{)}^{5x}(\frac{5}{\mathrm{log}2x}+5\mathrm{log}(\mathrm{log}2x)).$$

The only additional step that was done was to put the fractions over a common denominator of $\mathrm{log}2x$ and factoring:

$$\frac{dy}{dx}=5(\mathrm{log}2x{)}^{5x-1}(1+(\mathrm{log}2x)\mathrm{log}(\mathrm{log}2x)).$$

asked 2022-10-29

Determining the value of parameters given constraints

If

$$\frac{x(y+z-x)}{\mathrm{log}x}=\frac{y(z+x-y)}{\mathrm{log}y}=\frac{z(x+y-z)}{\mathrm{log}z}$$

and

$$a{x}^{y}{y}^{x}=b{y}^{z}{z}^{y}=c{z}^{x}{z}^{y}$$

then what is the value of $a+\frac{b}{c}$?

I am getting as $a{x}^{y}{y}^{x}=b{y}^{z}{z}^{y}=c{z}^{x}{x}^{z}$ after solving the first equation but according to the question it is not correct. So now I have to proove only that ${x}^{z}={z}^{y}$, then my answer will be equal to $2$.

If

$$\frac{x(y+z-x)}{\mathrm{log}x}=\frac{y(z+x-y)}{\mathrm{log}y}=\frac{z(x+y-z)}{\mathrm{log}z}$$

and

$$a{x}^{y}{y}^{x}=b{y}^{z}{z}^{y}=c{z}^{x}{z}^{y}$$

then what is the value of $a+\frac{b}{c}$?

I am getting as $a{x}^{y}{y}^{x}=b{y}^{z}{z}^{y}=c{z}^{x}{x}^{z}$ after solving the first equation but according to the question it is not correct. So now I have to proove only that ${x}^{z}={z}^{y}$, then my answer will be equal to $2$.

asked 2022-09-06

solving equations with powers

Im trying to solve the equation

$3\cdot {2}^{-2/x}+2\cdot {9}^{-1/x}=5\cdot {6}^{-1/x}$

So far I tried applying logaritmas but it didnt prove helpful...are there any other ways?

Im trying to solve the equation

$3\cdot {2}^{-2/x}+2\cdot {9}^{-1/x}=5\cdot {6}^{-1/x}$

So far I tried applying logaritmas but it didnt prove helpful...are there any other ways?

asked 2022-09-18

Homework help to rearrange formula

Given the equation

${V}_{m}=u(\mathrm{ln}{m}_{0}-\mathrm{ln}{m}_{8})-g{t}_{f}$

I need to solve for ${m}_{0}$ Here is what I have but it looks messy and I feel like there is sometihng wrong or a better way

1st attempt

$\begin{array}{rl}& {V}_{m}=u(\mathrm{ln}{m}_{0}-\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}-{V}_{m}=0\\ & u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-{V}_{m}=g{t}_{f}\\ & u(\mathrm{ln}{m}_{0})=g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m}\\ & \mathrm{ln}{m}_{0}=(g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m})\xf7u\\ & {e}^{(g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m})\xf7u}={m}_{0}\end{array}$

2nd attempt - think this looks a little better but still not there yet

$\begin{array}{rl}& {V}_{m}=u(\mathrm{ln}\frac{{m}_{0}}{{m}_{8}})-g{t}_{f}\\ & {V}_{m}+g{t}_{f}=u(\mathrm{ln}\frac{{m}_{0}}{{m}_{8}})\\ & \frac{{V}_{m}+g{t}_{f}}{u}=\mathrm{ln}\frac{{m}_{0}}{{m}_{8}}\\ & {e}^{\frac{{V}_{m}+g{t}_{f}}{u}}=\frac{{m}_{0}}{{m}_{8}}\\ & {m}_{8}{e}^{\frac{{V}_{m}+g{t}_{f}}{u}}={m}_{0}\end{array}$

Given the equation

${V}_{m}=u(\mathrm{ln}{m}_{0}-\mathrm{ln}{m}_{8})-g{t}_{f}$

I need to solve for ${m}_{0}$ Here is what I have but it looks messy and I feel like there is sometihng wrong or a better way

1st attempt

$\begin{array}{rl}& {V}_{m}=u(\mathrm{ln}{m}_{0}-\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}-{V}_{m}=0\\ & u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-{V}_{m}=g{t}_{f}\\ & u(\mathrm{ln}{m}_{0})=g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m}\\ & \mathrm{ln}{m}_{0}=(g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m})\xf7u\\ & {e}^{(g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m})\xf7u}={m}_{0}\end{array}$

2nd attempt - think this looks a little better but still not there yet

$\begin{array}{rl}& {V}_{m}=u(\mathrm{ln}\frac{{m}_{0}}{{m}_{8}})-g{t}_{f}\\ & {V}_{m}+g{t}_{f}=u(\mathrm{ln}\frac{{m}_{0}}{{m}_{8}})\\ & \frac{{V}_{m}+g{t}_{f}}{u}=\mathrm{ln}\frac{{m}_{0}}{{m}_{8}}\\ & {e}^{\frac{{V}_{m}+g{t}_{f}}{u}}=\frac{{m}_{0}}{{m}_{8}}\\ & {m}_{8}{e}^{\frac{{V}_{m}+g{t}_{f}}{u}}={m}_{0}\end{array}$

asked 2022-09-23

Why does ${\mathrm{log}}_{4}32\ne {\mathrm{log}}_{4}(4\cdot 8)$

${\mathrm{log}}_{4}32=2.5$

If

${\mathrm{log}}_{a}(b\cdot c)={\mathrm{log}}_{a}b+{\mathrm{log}}_{a}c\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}};(a>0,b>0,c>0,a\ne 1)$

Then why does ${\mathrm{log}}_{4}32$ can't be ${\mathrm{log}}_{4}(4\cdot 8)={\mathrm{log}}_{4}4+{\mathrm{log}}_{4}8=1+2=3$?

${\mathrm{log}}_{4}32=2.5$

If

${\mathrm{log}}_{a}(b\cdot c)={\mathrm{log}}_{a}b+{\mathrm{log}}_{a}c\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}};(a>0,b>0,c>0,a\ne 1)$

Then why does ${\mathrm{log}}_{4}32$ can't be ${\mathrm{log}}_{4}(4\cdot 8)={\mathrm{log}}_{4}4+{\mathrm{log}}_{4}8=1+2=3$?

asked 2021-05-08

Solve $\left(\mathrm{log}\left(7x\right)\right)\mathrm{log}x=5$

asked 2022-07-17

Is there a property for log(n)/n?

I found a small exercise which I couldn't figure what to do, so I found a solution. Then I tried to understand it and everything went well until I got to this part:

$\frac{1}{8}=\frac{\mathrm{log}(n)}{n}$

Then it just skipped and say that the answer was $n=43$. I was wondering if there is some kind of property for $\mathrm{log}(n)/n$ I don't know about. Or otherwise, how was this solved?

EDIT: This is the exercise Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size n, insertion sort runs in 8n2 steps, while merge sort runs in 64n lg n steps. For which values of n does insertion sort beat merge sort?

And this was the solution given:

$8{n}^{2}=64n\mathrm{log}(n)$

${n}^{2}=8n\mathrm{log}(n)$

$n=8\mathrm{log}(n)$

$\frac{1}{8}=\frac{log(n)}{n}$

I found a small exercise which I couldn't figure what to do, so I found a solution. Then I tried to understand it and everything went well until I got to this part:

$\frac{1}{8}=\frac{\mathrm{log}(n)}{n}$

Then it just skipped and say that the answer was $n=43$. I was wondering if there is some kind of property for $\mathrm{log}(n)/n$ I don't know about. Or otherwise, how was this solved?

EDIT: This is the exercise Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size n, insertion sort runs in 8n2 steps, while merge sort runs in 64n lg n steps. For which values of n does insertion sort beat merge sort?

And this was the solution given:

$8{n}^{2}=64n\mathrm{log}(n)$

${n}^{2}=8n\mathrm{log}(n)$

$n=8\mathrm{log}(n)$

$\frac{1}{8}=\frac{log(n)}{n}$

asked 2022-04-01

How to solve

$5-{\mathrm{log}}_{2}(x-3)={\mathrm{log}}_{2}(x+1)$