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.
Can all logarithm problems be solved algebraically?
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0@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... – 2012-02-10
2 Answers
$ \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.
$\log_2 (x-1)=\log_3(x+1) \Rightarrow \frac{\ln (x-1)}{\ln 2}=\frac{\ln (x+1)}{\ln 3} \Rightarrow \ln 3 \cdot \ln (x-1)= \ln 2 \cdot \ln (x+1) \Rightarrow$
$\Rightarrow \ln (x-1)^{\ln 3}-\ln (x+1)^{\ln 2}=0 \Rightarrow \frac{(x-1)^{\ln 3}}{{(x+1)^{\ln 2}}}=1 \Rightarrow (x-1)^{\ln 3}=(x+1)^{\ln 2} \Rightarrow$
$\Rightarrow (x-1)=(x+1)^{\log_3 2} \Rightarrow (x+1)-2=(x+1)^{\log_3 2}$
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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0You 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. – 2012-08-02