Backstory: I was watching a YouTube video by GradeAUnderA in which a formula appears. He claimed to have discovered it just plugging numbers in and just randomly proving things, so I decided to go ahead and look on the Internet for some proof to this formula. The only thing I found was a thread on a forum saying that it can't be true and that he is arrogant and enjoys drama.

I'm not interested in the drama, but in the maths. The formula is the following.

$$ x^n + y^m = x^{n + \frac{log(1+x^{\frac{m log y}{logx} -n})}{log x}} $$

Screenshot can be found here

So, I went ahead and decided to prove the formula, either wrong or right.

$$log_x(x^n + y^m) = n + \frac{log(1+x^{\frac{m log y}{logx} -n})}{log x} \tag{1}$$

$$log_x(x^n + y^m) - n = \frac{log(1+x^{\frac{m log y}{logx} -n})}{log x} \tag{2}$$

$$log_x(x^n + y^m) - log_x(x^n) = \frac{log(1+x^{\frac{m log y}{logx} -n})}{log x} \tag{3}$$

$$log_x(x^n + y^m) - log_x(x^n) = log_x(\frac{x^n + y^m}{x^n}) \tag{4}$$

$$ log_x(\frac{x^n + y^m}{x^n}) = log_x(1 + \frac{y^m}{x^n}) \tag{5}$$

$$ log_x(1 + \frac{y^m}{x^n}) = \frac{log(1+\frac{y^m}{x^n})}{log x} \tag{6}$$

$$ \frac{log(1+\frac{y^m}{x^n})}{log x} = \frac{log(1+x^{\frac{m log y}{logx} -n})}{log x} \tag{7}$$

$$ \frac{y^m}{x^n} = x^{\frac{m log y}{logx} -n} \tag{8}$$

$$ log_x(\frac{y^m}{x^n}) = \frac{m logy}{logx} -n \tag{9}$$

$$ log_x y^m - n = \frac{m logy}{logx} -n \tag{10}$$

$$ log_x y^m = \frac{m logy}{logx} \tag{11}$$

$$ \frac{log y^m}{log x} = \frac{m logy}{logx} \tag{12}$$

$$ \frac{m logy}{logx} = \frac{m logy}{logx} \tag{13}$$

Ok, so that's the whole proof. I'm not sure if I made any mistakes, but I'd say I did not. Can anybody correct me if I am wrong? Thank you very much.

  • $\begingroup$ As you obviously know how to typeset mathematics on this site, please also replace that ugly screenshot with a nice typeset formula. $\endgroup$ Aug 14, 2016 at 11:11
  • $\begingroup$ I put the screenshot there just to show that it appeared in a video. Will add it now. $\endgroup$ Aug 14, 2016 at 11:12
  • 1
    $\begingroup$ It cannot be true for all $x,y\in\mathbb{R}$ because it involves $\ln x$ and $\ln y$ (including division by $\ln x$). The RHS is not defined for $x=y=0$ and $x=1$ for example $\endgroup$
    – smcc
    Aug 14, 2016 at 11:13
  • $\begingroup$ @smcc Yes, that I figured out, I just wanted to know if the formula made any sense at all. $\endgroup$ Aug 14, 2016 at 11:17
  • $\begingroup$ @FrenzyLi will edit now, thanks $\endgroup$ Aug 14, 2016 at 11:17

1 Answer 1


Your proof is correct. Let me try to write it in a short way. We have that $$A:=\frac{\log(1+x^{\frac{m \log y}{\log x} -n})}{\log x}=\log_x(1+x^{\log_x y^m-n})=\log_x\left(1+\frac{y^m}{x^n}\right).$$ Hence $$x^{n+A}=x^n\cdot x^A=x^n\cdot \left(1+\frac{y^m}{x^n}\right)=x^n+y^m.$$

P.S. Note that the formula holds if $x>0$, $x\not=1$, and $y>0$.

  • $\begingroup$ @smcc Thanks a lot! $\endgroup$
    – Robert Z
    Aug 14, 2016 at 11:19
  • $\begingroup$ Thank you every much. I supposed I went the long way because I didn't want to make any mistake. $\endgroup$ Aug 14, 2016 at 11:19
  • $\begingroup$ @ChemiCalChems Yours is a wise approach! $\endgroup$
    – Robert Z
    Aug 14, 2016 at 11:22
  • $\begingroup$ I just have one question. Logarithms of negative numbers give complex values don't they? Maybe this is defined for and real numbers that is not 0 or 1. $\endgroup$ Aug 14, 2016 at 11:23
  • $\begingroup$ @ChemiCalChems Taking the principal value $\text{Log} z$ it should work also for negative numbers. $\endgroup$
    – Robert Z
    Aug 14, 2016 at 11:52

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