# Proof $\log_b a\cdot\log_c b\cdot\log_a c=1$,

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.

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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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Your 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 ;) – Clive Newstead Sep 27 '12 at 14:17
Here's how the last bit should have been written: $\log_b a\cdot\log_c b\cdot\log_a c$ $=\log_c a\cdot\log_a c$ $=\dfrac{1}{\log_a c}\cdot\log_a c = 1$. – Michael Hardy Sep 27 '12 at 18:05

${\bf Hint}\quad\begin{array}{cccccc} &\rm x^{\,I} &\rm C\quad\\ & \ \nearrow & \\ \rm A\!\!\!\! & & \downarrow \rm x^{\,J} \\ & \nwarrow & \\ &\rm x^K &\rm B\quad\ \ \end{array}\rm\ \Rightarrow\ \ IJK\, =\, 1$

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That's an interesting way to put it. – Snowball Oct 4 '12 at 22:05

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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See my answer for a more graphic viewpoint. – Bill Dubuque Oct 4 '12 at 22:15

By definition $\log_a b = \frac{\log b}{\log a}$.

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This is not true by definition. It's a provable result. – Michael Hardy Sep 27 '12 at 18:03
Good point. I forget to think of it that way. – Karolis Juodelė Sep 27 '12 at 18:08

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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... or to a logarithm of arbitrary base as long as it's the same one everywhere. – Henning Makholm Sep 27 '12 at 16:14