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.
[Math] Proof $\log_b a\cdot\log_c b\cdot\log_a c=1$,
logarithms
Related Solutions
The last line is simply the application of the first two: $$\log_b(x) = \frac{\ln(x)}{\ln(b)}= \ln(x) \cdot \frac{1}{\ln(b)} = \frac{\ln(x)}{\ln(a)} \cdot \ln(a) \cdot \frac{1}{\ln(b)} = \frac{\ln(x)}{\ln(a)} \cdot \frac{\ln(a) }{\ln(b)} = \log_a(x) \cdot \frac{\ln(a) }{\ln(b)}$$
Since $a$ and $b$ are constants, $\frac{\ln(a) }{\ln(b)}$ is constant as well.
If you set $X=\log a$, $Y=\log b$ and $Z=\log c$ (any base you like), your condition becomes easily $$ \frac{X^3+Y^3+Z^3}{XYZ}=3 $$ Now you can use the identity $$ X^3+Y^3+Z^3=(X+Y+Z)^3-3(X+Y+Z)(XY+YZ+ZX)+3XYZ $$ to get that $$ (X+Y+Z)\bigl((X+Y+Z)^2-3(XY+YZ+ZX)\bigr)=0 $$ The second factor can be written $$ X^2+Y^2+Z^2-XY-YZ-ZX $$ and we should consider the quadratic form having as matrix $$ A=\begin{bmatrix} 1 & -1/2 & -1/2 \\ -1/2 & 1 & -1/2 \\ -1/2 & -1/2 & 1 \end{bmatrix} $$ Since $$ \det[1]=1>0, \qquad \det\begin{bmatrix}1 & -1/2 \\ -1/2 & 1\end{bmatrix}=3/4>0 \qquad \det A=0 $$ the quadratic form is positive semidefinite. Its null space contains the vector $[1\;1\;1]^T$, so we can conclude that your condition implies $$ X+Y+Z=0\qquad\text{or}\qquad X=Y=Z $$ Since by assumption $a$, $b$ and $c$ are pairwise distinct, we end with $X+Y+Z=0$.
Without quadratic forms, you can reason about $X^2+Y^2+Z^2-XY-YZ-ZX=0$ as follows. Since $Z\ne0$ by assumption, we can set $X=uZ$ and $Y=vZ$, so the equation becomes $$ u^2-uv+v^2-u-v+1=0 $$ and, solving with respect to $v$, $v^2-v(u+1)+u^2-u+1=0$, the discriminant is $$ (u+1)^2-4(u^2-u+1)=-3(u-1)^2 $$ so the equation has a solution only for $u=1$, which gives $v=1$. Therefore $X=Y=Z$.
Another “elementary” approach. Suppose $(X+Y+Z)^2=3(XY+YZ+ZX)$. Set $s=X+Y+Z$ and $p=XYZ$; then $X$, $Y$ and $Z$ are the roots of the equation $$ t^3-st^2+\frac{s^2}{3}t-p=0 $$ by Viète's formulas. We can complete the cube getting $$ \left(t-\frac{s}{3}\right)^3=p-\frac{s^3}{27} $$ which should have three real roots. This is impossible unless the roots are coincident, so $p=s^3/27$ and $X=Y=Z$.
Best Answer
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.