A Nice Exponential Equation from SyberMath 😉

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[@moeberry8226]

It’s actually x=(2n+1)ipi )/(ln(3)+2ipik) where k is also any integer.

[@krishnamoyghosh6047]

Classic problem. It demands so deep thoughts. Kudos to u. Please keep it up.

[@SyberMath]

Thank you! Will do

[@mehrdadbasiri9968]

Beautiful one...👌👌👌.

[@SyberMath]

Thanks!

[@scottleung9587]

Nice - I also got the same family of solutions! I love all these trips to the complex world.

[@vladislavlukmanov4998]

If u had divided by 6^t u would have gotten (5/2)^t+(5/3)^t=1. On the left hand side is an increasing function, on the right is a constant, so it means that the equation has 1 solution at most, and u just might guess t=-1. It is simpler than u presented.

[@SyberMath]

Totally! Thanks for sharing

[@leif1075]

@@SyberMath you could could take the ln on both sides at 8:00 right since you can take complex log..and indid forget about the 2n multiple..doesn't that happen to you..thanks for sharing.

[@francisco-kb7mv]

The mathematics, I love.

[@ochukoobrikogho5587]

Wow! I'm amazed. Great job!

[@SyberMath]

Thank you

[@wesleydeng71]

1/15 + 1/10 = 1/6 is obvious.

[@francisco-kb7mv]

The solution is nice.

[@RexxSchneider]

It's always useful to get a "feel" for what a solution might look like. Obviously we can simplify by writing u = 3^x, and then we know that 15^u and 10^u increase faster than 6^u as u increases. Since the LHS is already bigger than the RHS when u = 0 ( 1+1 > 1), it should become clear that u has to be negative. Trying u = -1 immediately gives a solution (1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6).
Values of u < -1 will always make the LHS smaller than the RHS and values of u > -1 will always make the LHS bigger than the RHS, because of the rate of change of the exponentials, so that's our sole solution. Of course, we still have to find x = ln(u)/ln(3) = ln(-1)/ln(3) = ln(e^(2nπi + πi) / ln(3) = (2nπi + πi)/ln(3) -- where n is any integer.

[@goldfing5898]

First, you should substitute 3^x = t:
15^t + 10^t = 6^t
We search for integer solutions first. t = 0 is no solution, since
15^0 + 10^9 = 1 + 1 = 2 > 1 = 6^0.
And for every positive t, 15^t + 10^t is greater than 6^t,
e.g. for t = 1, 15 + 10 = 25 > 6, and these exponential functions are monotonically increasing.
So t must be negative. An easy to find solution is t = -1, since
15^(-1) + 10^(-1) = 1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6 = 6^(-1).
But:
3^x is positive for any real x, so t = 3^x can actually not be negative...
So there is no real solution (?)

[@RexxSchneider]

Indeed, but there are complex solutions if you write -1 = e^(2nπi + πi), where n is any integer. Then you can take natural logs of 3^x = -1 to get all the values for x.

[@michaelcolbourn6719]

Maybe a silly question, but you've given the solution for variable x involving a new unknown variable of n, so how have we solved for x of we don't know n?

[@maigretus1]

But we do know what n is. n is any non-negative integer (0, 1, 2, 3...). There are an infinite number of solutions.

[@michaelcolbourn6719]

@@maigretus1 gotcha. Thank you.

[@moeberry8226]

@@maigretus1n can be any integer not just nonnegative. n can be -1 for example.

[@barakathaider6333]

👍

[@2012tulio]

3^X = -1
After that , No real solutions ..

[@mdatik5517]

Help me for solving the exponential equation: y=2^{(y-2)÷2}. Some problems also for the commenters:(3x+1)(4x+1)(6x+1)(12x+1)=5, x^7+1÷x^7=91√7,x^3-1÷x=4,(x^2-2√2x)(x^2-2)=2021,4÷(x^2+1)^2-1÷(x^2-1)^2=1÷4x^2,√(2020+√x)-x=20,(x^3+3x^2-4)^1÷3-x=(x^3-3x+2)^1÷3,(x^3-3x)^2+(x^2-2)^2=4

[@rakenzarnsworld2]

2/15 + 3/15 = 1/6
x = -1/3

[@SyberMath]

Are you sure?

[@broytingaravsol]

x=i(1±2n)π/ln3

[@moeberry8226]

There’s no need for plus or minus since n is any integer.

[@broytingaravsol]

@@moeberry8226 got it

[@GAMINGTV-tj3on]

X=(iπ+°°)/ln3 and this

[@true7781]

-"X=(iπ+°°)/ln3 and this"-
𝒙 = 𝝅𝒊(1 + 2𝒌)/𝑳𝒏 3 =
𝝅𝒊(1 + 2𝒌)/(𝒍𝒏 3 + 2𝝅𝒍𝒊) =
𝝅𝒊(1 + 2𝒌)(𝒍𝒏 3 - 2𝝅𝒍𝒊)/[(𝒍𝒏 3)² + (2𝝅𝒍)²] =
𝝅(1 + 2𝒌)(2𝝅𝒍 + 𝒊(𝒍𝒏 3))/[(𝒍𝒏 3)² + (2𝝅𝒍)²], 𝒊² = - 1, 𝒌,𝒍 ∈ℤ

[@giuseppemalaguti435]

No real solution