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Solve this equation $$\displaystyle \log_2x=\log_{5-x}3$$

the answer is $x=2,x=3$

Can you give me some hint

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Since the function of the left hand side is concave and the function of the right hand side is convex, the equation has at most two solutions. – medicu Jun 4 '13 at 11:51
@medicu for real x's only, right? – Simple Art Jan 6 at 21:45

It’s certainly at least worth trying to see what value of $x$ makes the arguments of the logs equal and what value makes the bases equal, and when it turns out that the same value, $x=3$, does both, you’re done: the two expressions are then identical.

Added: Having spotted one trivial solution, it’s worth thinking about whether there are others. Are there any values of $x$ that make either side especially simple? If $x=2$, the lefthand side is $\log_22$, which we can immediately evaluate exactly as $1$. And pleasantly enough, it just happens that the righthand side is then $\log_33$, which of course is also $1$.

A generally solution is hard. If you have the equation $\log_b x=\log_{a-x}c$, you can reduce both sides to the same base. If you pick natural logs, you use the fact that $$\log_bx=\frac{\ln x}{\ln b}$$ to write $$\frac{\ln x}{\ln b}=\frac{\ln c}{\ln(a-x)}\;,$$ or $\ln x\ln(a-x)=\ln b\ln c$. This is very ugly in general, but note that $x+(a-x)=a$; if you’re lucky enough to find that $b+c=a$, as in this problem, you clearly get two solutions by setting $x=b$ and $x=c$.

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You are just randomly observing that $x=3$ is a solution, it does not explain why nor does it give the other solution... – Myself Jun 4 '13 at 3:43
@Myself: You’re mistaken: I clearly explained the reasoning, and the procedure is obviously not random. – Brian M. Scott Jun 4 '13 at 3:45
How is it different from solving $x^2+x = x^3 + x^5$ by trying to find a solution with $x^2=x^3$ and $x = x^5$? My point being that there is no reason to expect that the solution will satisfy an additional contraint. – Myself Jun 4 '13 at 3:50
Ok, sorry, I realise that sounded like a harsh comment for what is a good answer. My comment should merely be thought of as a warning that the sketched 'method' is not really a 'method' that will produce a solution in any case. (Two additions might be the to use some calculus to show that these are the only solutions and to give a 'general' solution in terms of the LambertW-function, if it can be done.) – Myself Jun 4 '13 at 4:11
@Myself: Oh, I quite agree that there’s no reason to expect the kind of good fortune that we get here. But it never hurts to try some simple-minded things first, if only to get a better feel for what’s going on, and it’s certainly better to do so than to sit there ‘admiring’ the difficulty of the problem! – Brian M. Scott Jun 4 '13 at 4:13


$$\implies \frac{\log x}{\log2} = \frac{\log3}{\log (5-x)} $$

$$ \implies \log x \cdot \log(5-x) = \log2 \cdot \log3$$

$$\implies \log\log(5-x)^x = \log\log(3)^2$$

or $$\log\log(5-x)^x =\log\log(2)^3$$

Comparing both, we get either $x = 2$ , or $x = 3$.

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$\ln \ln 3^2=\ln(2\ln 3)=\ln 2+\ln\ln 3\ne(\ln 2)(\ln 3)$ – Brian M. Scott Jun 4 '13 at 3:54
I was gonna write almost a same answer until i read yours. But i fail to understand how did you reach the third implication. So im writing my thing. – Rohinb97 Jun 9 '13 at 7:05

By the above equation,


Now x must be between 5 and 0. Seeing some logical cases, 5-x is 3 or x is 3. Hence you get the answers above. Do i have to prove there is no other complex x for which this equation holds true?

I do not think there will be a complex answer, since x can only lie between 0 and 5. One thing we should try is log2/log(5-x)=logx/log3, but i do not thing it will yield anything. Since log2*log3 is constant, the set of possible x should be very limited.

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