Solving an equation

[past schooner]

I have taken logarithms in both sides but I don't know what to do next

[proven sierra]

With logarithms, you can lower the multiplication into outside addition, and power $\log(a^b)$ to $b\log(a)$

[hardy sparrow]

.boomme

[proven sierra]

So, for instance, $\log((6-2x)^2 e^{2x-6})$ would be identical to $(\log((6-2x)^2) + \log(e^{2x-6})$

[hardy sparrow]

.boomme

[proven sierra]

Then, $\log((6-2x^2))$ is $2\log(6-2x)$

[hardy sparrow]

.boomme

[proven sierra]

Lastly, don't forget: $\log e=1$, and $e^{\log x}=x$

[hardy sparrow]

.boomme

[proven sierra]

I'll be here for any qs

[past schooner]

Even if u get logarithms

U can't solve it

I have made in form $2\ln 2-2\ln (6-2x) +10-4x-2e^{2-x}=0$

[hardy sparrow]

~ 🎭 𝕩𝓡𝔦ｓ𝕋𝓐ŘÃş 🎭 ~

[past schooner]

Check if I did any mistake @proven sierra

[proven sierra]

yeah i agree that's hard, but try starting from here instead

in the last step, actually, try considering both pos and neg as necessary

also you can try the substitution $y=x-3$ and thus $-1-y=2-x$

[hardy sparrow]

.boomme

[proven sierra]

ok, here's what i've got so far. you do a sub with $y=x-3$, then solve for two solution sets: (1) when $e^{-1-y-e^{-1-y}}=ye^y$ and (2) when $e^{-1-y-e^{-1-y}}=-ye^y$. one unique solution for (1) exists near $y\approx 0.18$, but it cannot be obtained analytically. two solutions for (2) exist at $y=-1$ and $y=0$, but $y=0$ is extraneous. taking $y=-1$, you obtain an integer solution $x=2$ by re-substitution. in total, you will report two solutions: one near $x\approx 3.18$, and one at $x=2$. hope this helps

[hardy sparrow]

.boomme

[proven sierra]

i can't imagine this being a school problem!

^ also, why (1) can't be obtained analytically. you will be required to solve an equation where $e^{\mathrm{something}}$ is equal to $\ln{(\mathrm{something})}$, and in general only numerical solutions are possible. integral solutions are found by graphing, guessing, and then checking.

[hardy sparrow]

.boomme

[proven sierra]

elaboration (integral solutions). integral means relating to integers. so the integer solutions $y=-1$ and $y=0$. you get this by graphing. with a strong guess, you plug those in, and prove they are the solution.

[hardy sparrow]

.boomme

[past schooner]

How did you find the x=2?

@proven sierra

[proven sierra]

graphed, figured 2 is a possible answer. plugged in

[past schooner]

But without graph it's impossible to find it?

[proven sierra]

I was not able to find it. it ultimately asks you to solve something like e^??? = ln(???). those things are often solved numerically

[hardy pier]

boome

[past schooner]

What numerically means

[proven sierra]

?

[hardy pier]

after ur done can u help me abit

[past schooner]

An extra information is

U have to solve this equation in [0, 2]

[proven sierra]

it means, guess and check using by looking at numbers. perhaps by graphing, or a large enough table

[past schooner]

Okay

Can I tell you another equation?

[proven sierra]

2 is included in [0,2]. so your eq is solved. ✅

[proven sierra]

ok

[proven sierra]

ok

[past schooner]

$π+(2x-π) * \cosx - 2 \sin x =0$

[hardy sparrow]

~ 🎭 𝕩𝓡𝔦ｓ𝕋𝓐ŘÃş 🎭 ~
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[past schooner]

Damn

[proven sierra]

ok, $\pi+(2x-\pi)\cos{x}-2\sin{x}=0$

[hardy sparrow]

.boomme

[past schooner]

This one

[proven sierra]

well that's another tuff one, my hunch is to try one of those trig identities. i have forgotten all of it but i can use the internet. i would look up half angle and power formulas as keywords

[past schooner]

Do u know how somebody solved it?

[proven sierra]

elaborate. has someone solved it?

[past schooner]

Took it's derivative found the absolute minimum and maximum, whose y value was zero

Try it

[proven sierra]

are you saying, a related max/min finding problem can help you solve this problem?

i doubt that, sorry to disappoint u

however, notice

[past schooner]

Since u take it's derivative

[proven sierra]

(thinking)

[past schooner]

U can find critical points

Where f'(x)=0

[proven sierra]

yes

[past schooner]

He found them, and when he found f(min) and f(max) it was 0

[proven sierra]

that helps you find the peaks and the velleys

[past schooner]

So found and it's solutions

[proven sierra]

oh

i see

interesting way to arrive at the answer then

let me try that

[past schooner]

Okay

[proven sierra]

there are infinitely many zeros, and most of them do not coincide with any peaks or valleys. two of them do however

so i figure your friend has found x=0 and x=pi as the solutions. those are valid answers, but not complete

ultimately, you need to solve an eq of the form $x\cos{x}=\sin{x}+c$. that's also best solved numerically

[hardy sparrow]

.boomme

[past schooner]

Oh

I forgot to mention

$x \in [0, π]$

[hardy sparrow]

~ 🎭 𝕩𝓡𝔦ｓ𝕋𝓐ŘÃş 🎭 ~

[past schooner]

@proven sierra

[proven sierra]

Ok, well then you've got it

had to know that the minima are aligned with the zeros beforehand

[past schooner]

So the solutions are?

[proven sierra]

and graphing would have helped you w that

$x=0,\pi$

[hardy sparrow]

.boomme

[past schooner]

The goal is without graph

Now I need a help with a limit

[proven sierra]

well then. cos and sin are each either 0 and 1 at those intervals etc. so i would base my thought on that

*between

[past schooner]

$\lim_{{x \to \frac{\pi}{2}^+}} \frac{\pi x}{\pi \sin x - 2x}$

[hardy sparrow]

~ 🎭 𝕩𝓡𝔦ｓ𝕋𝓐ŘÃş 🎭 ~

[proven sierra]

Oh sorry went afk

ok let me look

alright

graph $\pi\sin x$ then argue that it acts like a constant function $y=\pi$ at around $x=\pi/2$

[hardy sparrow]

.boomme

[proven sierra]

Now replace the "$\pi\sin x$" with "$\pi$" in the denominator

[hardy sparrow]

.boomme

[proven sierra]

then it should be easy as showing that a polynomial division has uncanceled factors

ans: $-\infty$