How can you solve problems like $ x^{x-1}=7 $? More generally, how can you solve equations like $(ax+b)^{cx+d}=e$ , where $a,b,c,d,e$ are given?$($Give all the roots, including complex ones$)$
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2I think you are looking for the Lambert W-Function. Concerning your given equation it seems like even the Lambert W-Function does not help in order to produce a closed-form hence WolframAlpha only gives two numerical solutions. – mrtaurho Jan 04 '19 at 20:34
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I think WolframAlpha is only giving real solutions, but I want the complex solutions,too. – Math Lover Jan 04 '19 at 20:51
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I can understand you doubts but take a look at this similiar request which indeed produces complex values too $($take for example the $W_{-1}$ solution for $n=0$ which is a complex number, to be precise look here$. – mrtaurho Jan 04 '19 at 20:56
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Interesting enough for me to post an answer is the fact that the problem for your first equation only lies in the fact that the exponent is given by $x-1$ and not by $x$. I will demnonstrate how to solve this equation for the latter case. First of all rewrite the $x^x$ term in terms of the exponential and do not forget assuming a complex valued logarithm to get
$$\begin{align*} x^x&=7\\ e^{x\log(x)+2\pi i n}&=7\\ x\log(x)+2\pi i n&=\log(7)\\ \log(x)e^{\log(x)}&=\log(7)-2\pi i n\\ \log(x)&=W(\log(7)-2\pi i n) \end{align*}$$
$$\therefore~x=e^{W(\log(7)-2\pi i n)}~~~n\in\mathbb Z$$
I have doubts that on can deduce a general formula for arbitrary $a,b,c,d,e$ $($just take $a=c=1$,$b=0$,$d=-1$ and $e=7$ to reproduce your first equation$)$. Anyway considering that $a=c$ and $b=d$ it is indeed possible since this is basically the same as $x^x$ and can be solved using the Lamber W-Function but note that you have to consider the different branches of this function with regard to the values of $a,b$ and $e$.
From hereon I have to admit that I have not enough experience with the Lambert W-Function to give an detailed outline of the different branches and why they are important.
mrtaurho
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Can you explain a bitthe Lambert W-Function, and how you solved $x^x=7$? I don’t understand your steps. Thanks! Also, will it be helpful to multiply both sides of the equation by x, to get $x^x=7x$? – Math Lover Jan 04 '19 at 21:15
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The Lambet W-Function is defined as the inverse function $f(z)=ze^z$, i.e. $W(z)=f^{-1}(z)$. Thus, by having an equation of this structure on can apply this definition, namely $$ze^z=r\Rightarrow z=W(r)$$ By performing the steps to get this structure with your given equation leaves us with $$\log(x)e^{\log(x)}=\log(7)-2\pi i n\Rightarrow \log(x)=W(\log(7)-2\pi i n)$$Concerning the extra $2\pi i n$: this happens since the logarithm as a complex-valued function is far away from being injective. In fact the logarithm inherits a additive period from the exponential since $e^{2\pi in}=1$. – mrtaurho Jan 04 '19 at 21:34
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Since you were asking for the complex roots of your equation too I assume you are used to complex numbers in general. Thus, recall Euler's Formula $e^{it}=\cos(t)+i\sin(t)$ We know that the sine function aswell as the cosine function are both periodic with period $2\pi$. So plugging in $t+2\pi$ does not change the value $$e^{i(t+2\pi)}=e^{it}$$ And this can be taken further so that we can conclude $$e^{i(t+2\pi n)}=e^{it}~~~n\in\mathbb Z$$ This is what this extra term is all about: ensuring that we get all possible values of the exponential, and the logarithm repectively, at not just one. – mrtaurho Jan 04 '19 at 21:56
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Shouldn’t the right side of your third line of your solution be ln(7),instead of log(7)? – Math Lover Jan 04 '19 at 22:57
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Also, how did you get from the first line to the second line? Thanks – Math Lover Jan 04 '19 at 23:00
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Oh, I did not clarified it: I am used to use $\log(x)$ for the natural logarithm. Concerning your second question $$x^x=e^{x\log(x)}=e^{x\log(x)}\cdot1=e^{x\log(x)}\cdot e^{2\pi i n}=e^{x\log(x)+2\pi i n}$$ – mrtaurho Jan 04 '19 at 23:12
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Oh, I misunderstood you. Short answer: Combining thw basic logarithmic properties, namely $e^{\log(x)}=x$ and $\log(a^b)=b\log(a)$, we get $$e^{x\log(x)}=e^{\log(x^x)}=x^x$$ For a longer and more detailed answer take a look at this recent post dealing with this issue $($especially $\text{KM}101$'s answer might be of help$)$. – mrtaurho Jan 04 '19 at 23:36
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I just discovered something. If $e^{x log(x)}=x^x$, then $(e^x)(e^{log(x)})=x^x$,$(e^x)(x)=x^x$, so $e^x=x^{x-1}$, so problem I gave as a sample becomes $e^x*e^{2in pi}=7$,x=ln(7). Whoa!!!!! – Math Lover Jan 05 '19 at 00:25
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I just discovered something. If $e^{x log(x)}=x^x$, then $(e^x)(e^{log(x)})=x^x$,$(e^x)(x)=x^x$, so $e^x=x^{x-1}$, so problem I gave as a sample becomes $e^x*e^{2in pi}=7$,x=ln(7)-2pi in. Whoa!!!!! – Math Lover Jan 05 '19 at 00:31
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No, that is wrong. According to the power rules we got $x^a\cdot x^b=x^{a\color{red}{+}b}$ and not $x^a\cdot x^b\neq x^{a\color{red}{\cdot}b}$. Your conclusion is false aswell since overall you stated that $x^x=x^{x-1}$ which is certainly not true beside the case $x=0$. – mrtaurho Jan 05 '19 at 00:47
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No, I divided both sides by x, so it is correct besides x=0. x=0 isn’t an answer anyways. The left side was $e^x$ times x. – Math Lover Jan 05 '19 at 01:12
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As I already saided: your first step is sadly speaking completely wrong so there cannot be a right solution come out of this wrong initial step. The problem I see is the final "equality" stating that $e^x=x^{x-1}$. Just take $x=1$ so the whole thing becomes $e^1=1^{1-1}\Rightarrow e=1$. I guess you will agree with me that this is nonsense. I have to admit that I was wrong concerning the case $0$ as a solution but in fact $x^x=x^{x-1}$ only works out for $x=1$. – mrtaurho Jan 05 '19 at 01:17
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$$x^{x-1}=e^{(x-1)\log x}\implies x^{x-1}=7\iff e^{(x-1)\log x}=e^{\log 7}\implies(x-1)\log x=\log7$$
and you'll probably need some trascendental function like Lambert function or stuff.
DonAntonio
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