An Exponential Log

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[@ignaciodecastrofondevila2456]

And what is the solution if instead of 7 it is another constant?
How could we solve it?
(Thanks for the video)

[@Skank_and_Gutterboy]

I haven't tried it yet but I'll bet the problem is exponentially harder if the constant is other than 7.

[@stvp68]

Lol @ sometimes in the morning

[@Skank_and_Gutterboy]

After a minute of staring at this equation, what came to me was: Right side of the equation is a constant, 7, and on the left side we have expressions with 2 and 5. 2+5=7, so is there an x that yields 2+5? Yes, 10^log2 = 2 and 10^log5 = 5. x=10, bada-bing, done! You have to try things like this when dealing with a non-standard equation. I'll bet solving for x is A LOT harder if the right side was a different constant like 8 or 9.

[@ignaciodecastrofondevila2456]

With a few variable changes, the equation x^loga + x^log(b) = C comes out
Y+Y^r =C
with Y=10^[(log(a))*(log(x))] and r=log(b)/log(a)
On the left, a growing function, r whatever value it has (r not integer)
So there´s an unique real solution for Y (and so for x).
Maybe Newton's method... or maybe we are building a new Mandelbrot set... I propose to call it Sybertmath set.

[@trojanleo123]

x = 10

[@nasrullahhusnan2289]

@ignaciodecastrofondevilla2456:
Let consider a general equation [x^ln(a)]+[x^ln(b)]=c
where a, b, and c are constants
Noting x=e^ln(x) then
x^ln(a)=e^[x^ln(a)]
=[e^ln(a)]e^x
x^ln(b)=[e^ln(b)]e^x
Thus c=[e^ln(a)+e^ln(b)]e^x
=(a+b)e^x --> x=ln[c/(a+b)]

[@Skank_and_Gutterboy]

That derivation does not work.
Proof: assign a=2, b=5, c=7.
Then, x = ln[7/(2+5)] = ln(1) = 0.
Substituting this found value of x back into the original equation:
0^ln(2) + 0^ln(5) = 0 + 0 = 0. 0 does not equal 7, so x=ln[c/(a+b)] does not satisfy the problem equation.

[@nasrullahhusnan2289]

@@Skank_and_Gutterboy: Is there anything wrong in the outline I have proposed?

[@ignaciodecastrofondevila2456]

@@nasrullahhusnan2289 Yes, there is.
Fourth line, incorrect ------------> x^ln(a) = e^[x^ln(a)]
Fourth line, corrected -----------> x^ln(a)=e^ln([x^ln(a)])
And so on.
(BTW, video is with decimal logarithm)

[@Skank_and_Gutterboy]

@@nasrullahhusnan2289
Yes. Look at my post above where I prove that your solution for x does not satisfy the original problem equation.

[@nasrullahhusnan2289]

@@ignaciodecastrofondevila2456:
Thanks for your correction. But on your objection on the base of the logarithm we can easily change log(a) into ln(t): log(a)=ln(t) --> a=10^ln(t), t=e^log(a)