1

x = 24√3

i valori dell'equazione sono x = 1; x = 2/3

x = 1/4

Answer = 1 I used substitution because substitution is awesome!

x = 5^√2 or x = 5^(-√2)

ab/(a+b)=e , ab/(a+b)=e^2/e , ab=e^2 , a+b=e , b=(e+/-e*i*V3)/2 , a=(3e+/-e*i*V3)/2 ,

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Cos(90°-45°)+sin(15)=sin45°+sin15°=1+(√6-√2/4)

X-3=4(mod7) ???? I did not understand how you get this

X-3=4(mod7) ???? I did not understand how you get this

Not gonna lie, this is a smart way of doing it. I Would've turned cos15 into cos(45-30) and use the subtraction formula, and do the Same thing with sin.

12^(x-1/x) = 3 x-1/x = Log[12,3] x-1 = xLog[12,3] x(1-Log[12,3]) = 1 x(Log[12,12]-Log[12,3]) = 1 x(Log[12,4]) = 1 x = 1/Log[12,4] 8^x = 8^(1/Log[12,4]) = 12^(Log[12,8^1/Log[12/4]) = 12^(Log[12,8]/Log[12/4]) = 12^(3Log[12,2]/2Log[12/2]) = 12^(3/2) = 12√12 =24√3

Lol I need to stop doing the problems from the thumbnail. I got stuck cause I was like what are we solving? Then again, I should have known that there's 1 equation with 2 unknowns, so it's not like I could have solved for either

@ SyberMath Shorts -- You should put the arguments inside of grouping symbols: ln[(a + b)/3] = [ln(a) + ln(b)]/2.

Solve my problem if you are really capable of I guess IT IS AS FOLLOWS: if x^2+x=1 then x^7+34/x+1=?

Hey why aren't you solving my question ⁉️⁉️⁉️⁉️⁉️⁉️⁉️😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡😡

12^(x - 1) = 3^x 4^x = 12 4 = 8^(2/3) 8^(2x/3) = 12 *8^x = 12^(3/2)* = (24)3^(1/2)

Looking at t³ - 5t - 2 = 0 we see t = -2 is a solution so we can write it as (t + 2) (t² - 2t - 1) = 0. But t needs to be non-negative for x to be real so there is only one solution for t: √x = t = 1 + √2 and x = t² = 3 + 2√2 and finally x - 2√x = 3 + 2√2 - 2 - 2√2 = 1.

I got a= ln((rt5+1)/2)/ln(9/2) by sorting the original equation into powers of 2 and 3, then treating as a quadratic in 3^2a. This gave 3^2a = 2^a((1+rt5)/2), and then taking logs and collecting "a" terms gave the answer. Nice.

4^a+18^a=81^a a=Log([2/9],(Sqrt[5]-1)/2)]=Log[4.5,0.5Sqrt[5]+0.5]~0.3199382066980040650491657338758298789841474274307941035072922216553638740432712673676202882151022345568954512788314008295516897770574500895314558551810515108966196821912946789984249109970485757976272891247601161060854534045237954921775901870466452594708064040710825614969389311396725943810069627458873857740102762027328842371488327805191226529079309714983851383816360259523138089400499493212881856663337223518517806062164722169591715442120927184824912516460334001125961205039613348619927712018944903984958137803761773454915239078384365460429462672969437662355735692718486786264118248161461134876339828354010034648894128545740815025971274963228614763119163876221449672201714767052756142625562587746081088455193290152157018327792100795144338513785524625687468784816564943948552036642828837755751930630811930731113256613021884585717878594628658362044717189939360718556814169386968509072761899353240882373754190407800266190741843

Gracias por compartir el interesante video sobre inecuaciones algebraicas 😊❤😊.

Solution: 2^(x²)+4^(x²) = 8^(x²) |/2^(x²)≠0 ⟹ 1+2^(x²) = 4^(x²) ⟹ 1+2^(x²) = 2^(2*x²) |with u = 2^(x²) ⟹ 1+u = u² |-u-1 ⟹ u²-u-1 = 0 |p-q-formula ⟹ u1/2 = 1/2±√(1/4+1) = 1/2±√(5/4) = (1±√5)/2 ⟹ u1 = (1+√5)/2 and u2 = (1-√5)/2 < 0 ⟹ 1st case: 2^(x1²) = u1 = (1+√5)/2 |lb() ⟹ x1² = lb[(1+√5)/2] |√() ⟹ x11/2 = ±√{lb[(1+√5)/2]} ≈ ±0,8332 x11 ≈ +0,8332 x12 ≈ -0,8332 2nd case: 2^(x2²) = u2 = (1-√5)/2 < 0 [that is not defined]

1/x^(x)^1=2=1/16 <=> 1/4^(4)^1/2 <=> 1/4^2=1/16 <=> x=4, or 1/2^2^2=1/16<=> x=2^2

No sound?

Did you upload It without any sound?

This is wrong. The domain of 1-x includes -1 and ±i while the original expression is undefined for these values. Therefore they are not equivalent.

Step 1. Learn to recognize that problems like this are associated with the Golden Ratio, the positive root of Φ^2-Φ-1=0 or φ^2+φ-1=0. Step 2. Divide the given equation by one of its terms and rearrange the results into one of these two forms. Step 3. Note that Φ=(√5+1)/2, φ=(√5-1)/2, and Φφ=Φ-φ=1. Step 4. Try associating the given equation with each of these two forms and see why some use (√5+1)/2 and others use (√5-1)/2.

a=log(2/9)^((5^(1/2)-1)/2)

So 2a+1/a=2. So a=0.5+0.5 sqrt 3 and b=1+sqrt 3 or a=0.5-0.5 sqrt 3 and b=1-sqrt 3.

Need more of these vids in my feed. Need to get my brain thinking every now and then

We know √2 is 1.414, now 2^7 >127 instant soln

Interesante ejercicio de simplificación de radicales, gracias por compartir 😊. Simplificación por racionalización del radical del denominador. 😊❤😊.

Wooooowwwww

I can't with this guy. He has a main channel for videos, creates a secondary channel specifically for shorts and then uploads long videos on it. Why'd you even called it "SyberMath Shorts"?

il risultato è x = 9 , x = - 9

The denominator: X1=-5; X2=2

2nd method is much simpler. I always rewrite log (base b) of some number as ln (some number) / ln (base b). Expression becomes ln(x)/ln(5) = ln(25)/ln(x). Cross multiplying, simplifying ln(25) as 2 ln(5), and taking the square root gives the answer as x = 5^(+/- sqrt(2)).

nice video!! subscribed!

Interesante ejercicio de una ecuacion con valor absoluto, gracias por compartir 😊

log10->(1)-loge->(2)= -0,69 [1-0,69=0,31] log10->(0,3) -loge->(0,6)= -0,01 <=> log10->(0,29) - loge-> (0,58)=0,007 ; log10->(0,29)=-0,54; loge->(0,58)=-0,54; x=0,29

1/16=(¼)² Change the exponent: 2=1/½ =1/sqrt(¼) x=¼

Sorry, I don't get why √x ≠ -2. If x is negative then √x will be a complex number, but the converse isn't necessarily true. for √x = -2, x = 4. Shouldn't (√x)² - 2√x = (-2)² - 2(-2) = 8, also be a valid solution?

We need more shorts on Trigonometric equations

We need more like this trigonometry

[(a^3+b^3)/2]^2=[(a^3-b^3)/2]^2 + (ab)^3 where a & b are both even or both odd positive Integers.

All of it is very nice and useless at the same time, unless someone wants to play some nice math tricks and substitutions. Cubic equations are solvable from about 16th century, and especially in their canonical form. This equation is nothing special and moreover it is in its canonical form, and requires five easy calculations to arrive at three real roots of it.