# Can all logarithm problems be solved algebraically?

Trying to solve $\log_2(x-1)=\log_3(x+1)$ and can't seem to get it algebraically. Tried changing bases, moving things around, but can't seem to crack it.

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To answer the question in the title: nope. Even things as simple as $x\log\,x=3$ require the use of a different sort of function for expressing solutions. For your actual problem: barring a fair bit of cleverness, I see no straightforward solution... – J. M. Feb 10 '12 at 17:45
@J.M. are there any results which show that? Something similar to Abel–Ruffini theorem regarding algebraic solution of polynomial equation? – user2468 Feb 10 '12 at 18:30
The keyword here is transcendental equations. AFAIK there is no general algorithm to solve them. Only tricks to solve specific ones.. I can't seem to find any theorems on that though. – aelguindy Feb 10 '12 at 19:06
@J.D. Not sure about general, but the example I gave requires the Lambert function to express the solution. Only slightly more complicated transcendental equations don't even have the luxury of an "easy" closed form... – J. M. Feb 10 '12 at 22:51

$$\log_2(x-1)=\log_3(x+1)$$ $$\log_2(x-1) = \frac{\log_2 (x+1)}{\log_2 3}$$ $$(\log_2 3)(\log_2(x-1)) = \log_2 (x+1)$$ $$\log_2((x-1)^{\log_2 3}) = \log_2 (x+1)$$ $$(x-1)^{\log_2 3} = x+1$$ At this point I might apply Newton's method.

Pedja's earier answer did everything right until a mistake near the end, but I prefer not to bring in $e$ when the number $e$ is not essential to the situation. Avoiding $e$ was really the reason why I felt this is worth answering.

Later note: Pedja has fixed the error.

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$$\log_2 (x-1)=\log_3(x+1) \Rightarrow \frac{\ln (x-1)}{\ln 2}=\frac{\ln (x+1)}{\ln 3} \Rightarrow \ln 3 \cdot \ln (x-1)= \ln 2 \cdot \ln (x+1) \Rightarrow$$

$$\Rightarrow \ln (x-1)^{\ln 3}-\ln (x+1)^{\ln 2}=0 \Rightarrow \frac{(x-1)^{\ln 3}}{{(x+1)^{\ln 2}}}=1 \Rightarrow (x-1)^{\ln 3}=(x+1)^{\ln 2} \Rightarrow$$

$$\Rightarrow (x-1)=(x+1)^{\log_3 2} \Rightarrow (x+1)-2=(x+1)^{\log_3 2}$$

If we make substitution $~u=(x+1)~$ we get :

$u-u^{\log_3 2}=2$

According to WolframAlpha this equation can be solved using numerical methods , so:

$u \approx 4.6298 \Rightarrow x \approx 3.6298$

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Sorry: how do you go from $(x-1)^{\ln 3} = (x+1)^{\ln 2}$ to $(x-1) = (x+1)^{\ln(2/3)}$? I can see $(x-1)=(x+1)^{(\ln 2)/(\ln 3)}$, but not what you wrote... – Arturo Magidin Feb 10 '12 at 19:00
....and $\ln 2/\ln 3 = \log_3 2$. – Michael Hardy Feb 10 '12 at 19:05
@ArturoMagidin,right...it should be $\log_3 2$, fixed..thanks – pedja Feb 10 '12 at 19:09
+1 since the error was fixed. I posted an answer that some might view as essentially the same, but I explained why. – Michael Hardy Feb 10 '12 at 19:13
You can even avoid Wolfram Alpha and get a numerical answer yourself easily enough using an iterated system. Rewrite the final equation as $u = 2 + u^{\log_3{2}}$. Now, since the graphs $y=u$ and $y=2 + u^{\log_3{2}}$ intersect in only one point, and the derivative of the latter at that point is less than $1$ in absolute value, the iterated values $u_{n+1} = 2 + {u_n}^{\log_3{2}}$ will converge to the solution. Starting with $u_1 = 1$ leads to $u \approx 4.6298$ in a dozen steps. – Théophile Aug 2 '12 at 19:48