Help with math

>>182770
What am I solving for?

Am I simplifying, finding x?

>>182773
Finding x

>>182775
2.738 for the first 3 digits

>>182781
Thank you so much, Could you upload a photo of your method?

>>182770
Not sure if this is the most efficient way of doing it, but

5^(x+1) = 3^(2*x) [log5 both sides]
log5(5^x+1) = log5(3^2x)
x+1 = log3(3^2x)/log3(5) [change base on right side]
x+1 = 2x/log3(5)
(x+1)/2x = 1/log3(5)
x/2x +1/2x = 1/log3(5)
1/2 + 1/2x = 1/log3(5)
1/2x = 1/log3(5) - 1/2
1/x = 2(1/log3(5) - 1/2)
x = 1/(2(1/log3(5) - 1/2))
x = 2.738132741922803

>>182787
Last digits are probably floating point error, by the way.

>>182787
Thank you so much, I love you guy.

>>182792
You're welcome. Love you too.

>>182787
Thanks for saving me, i was halfway through typing that before I realized i was explaining it like shit I'm horrible at logarithms so once I figured out x I couldnt explain how I had gotten it. I tried to help, thanks for the save !

>>182795
No problem. I was almost done when you posted your answer, so I just went ahead and posted mine as well. :)

>>182797
>>182795
Guys like you are the reason why I love 4chan

>>182787 #
>x = 2.738132741922803
x = 1/(2(1/log3(5) - 1/2)) is the answer. People need to stop converting to decimal approximations for no reason, especially in pure math. Normally I don't give a shit about this but it's important to break this need to always write decimal approximations, especially for someone starting higher maths.

1/(2(1/log3(5) - 1/2)) is a number just like 5, e, π, root(2), or 29/75 is; there is no need to write its truncated decimal approximation.

It basically comes down to what a number is, numbers are not just defined by where they fall on the complex plane or the number line, they are defined by what they do. When you write 1/(2(1/log3(5) - 1/2)) I can tell it is a number that satisfies 5^(x+1)=3^(2x), that's something you don't get to see with a decimal approximation.

For example, 0.5671432904 is very close to a fairly interesting number. But me telling you that doesn't convey much information. If I instead say that it is close to W(1) where W satisfies W(x)*(e^(W(x)))=x then that gives much more information, it's even enough information to create methods to get arbitrary decimal approximations of W(1).