Please help me proof $\log_b a\cdot\log_c b\cdot\log_a c=1$, where $a,b,c$ positive number different for 1.
Proof $\log_b a\cdot\log_c b\cdot\log_a c=1$,
5 Answers
Before we prove the given identity proof this idenity
$$\log_b a\log_c b=\log_c a$$
Proof: Implement the formula $\log_a b=\frac{\log_x b}{\log_x a}$
$$\frac{\log a}{\log b}\cdot\frac{\log b}{\log c}=\frac{\log a}{\log c}=\log_c a$$
Now proof the given identity.
$$\log_b a\cdot\log_c b\cdot\log_a c=1$$
$$\log_c a\cdot\log_a c=1$$
$$\frac{1}{\log_a c}\cdot\log_a c=1$$
$$1=1$$
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2Your logic is flawed here. For instance, $3 = 0 \Rightarrow 0 \times 3 = 0 \times 0 \Rightarrow 0 = 0$ but that doesn't mean that $3=0$ is true. If anything, a proof would go backwards through what you wrote ;) – 2012-09-27
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0Here's how the last bit should have been written: $\log_b a\cdot\log_c b\cdot\log_a c$ $=\log_c a\cdot\log_a c$ $=\dfrac{1}{\log_a c}\cdot\log_a c = 1$. – 2012-09-27
Change all to the natural logarithm $\log\,$:
$$\log_ba\cdot\log_cb\cdot\log_ac=\frac{\log a}{\log b}\frac{\log b}{\log c}\frac{\log c}{\log a}$$
and voila.
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1... or to a logarithm of arbitrary base as long as it's the same one everywhere. – 2012-09-27
${\bf Hint}\quad\begin{array}{cccccc} &\rm x^{\,I} &\rm C\quad\\ & \ \nearrow & \\ \rm A\!\!\!\! & & \downarrow \rm x^{\,J} \\ & \nwarrow & \\ &\rm x^K &\rm B\quad\ \ \end{array}\rm\ \Rightarrow\ \ IJK\, =\, 1$
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1That's an interesting way to put it. – 2012-10-04
By definition $\log_a b = \frac{\log b}{\log a}$.
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1This is not true by definition. It's a provable result. – 2012-09-27
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0Good point. I forget to think of it that way. – 2012-09-27
Let $\log_b a=x\implies b^x=a,$
$ \log_c b=y\implies c^y=b$ and
$\log_a c=z\implies a^z=c$
Now, $a^z=c\implies (b^x)^z=c\implies ((c^y)^z)^x=c\implies c^{xyz}=c\implies xyz=1$ assuming $c\neq 0,1$
Thus, $xyz=1\implies \log_b a\cdot\log_c b\cdot \log_a c=1$
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0See my answer for a more graphic viewpoint. – 2012-10-04