Prove that $x^{18} = y^{21} = z^{28}$ If $3, 3\log_yx, 3\log_zy, 7\log_xz$ are in A.P, then prove that thing in the title.
I tried to equate: $3(\log_yx -1) = 3(\log_zy - \log_yx) = 7\log_xz - 3\log_zy$ . Will it be of any help?
 A: You can rewrite the logs as $\log_y x^3$, $\log_x y^3$, and $\log_x z^7$. The arithmetic progression gives you a number $t$ with the property that \begin{align*} \log_y x^3 &= 3 + t \\ \log_z y^3 & = 3 + 2t \\ \log_x z^7 &= 3 + 3t. \end{align*}
Rewrite in exponential form:
\begin{align*} x^3 &= y^{3+t}  \\ y^3 &= z^{3+2t}  \\ z^7 &= x^{3+3t}  \end{align*}
We can try to get a sense of what $t$ is doing: 
$$x^3 = y^{3+t} = (y^3)^{\frac{3+t}{3}} = (z^{3+2t})^{\frac{3+t}{3}} = (z^7)^{ \frac{3+2t}{7} \frac{3+t}{3}} = (x^{3+3t})^{ \frac{3+2t}{7} \frac{3+t}{3}} = (x^3)^{(1+t)\frac{3+2t}{7} \frac{3+t}{3}}. $$
This forces 
$$ (1+t)\cdot \frac{3+2t}{7} \cdot  \frac{3+t}{3} = 1$$
whose only real solution is $t = \frac 12$.
The first two equations above become $x^3 = y^{7/2}$ and $y^3 = z^4$ which lead immediately to $x^{18} = y^{21}$ and $y^{21} = z^{28}$ as desired.
A: Hint:
Let $\log_xz=a\implies z=x^a$ and $\log_yx=b\implies x=y^b,z=y^{ab}$
$$\implies3,3b,\dfrac3{ab},7a$$ are in AP
If $3b-3=d,7a=3+3d\implies b=?,a=?$
$\dfrac3{ab}=3+2d\iff(3+d)(3+2d)(3+3d)=63$
One of the values of $d$ is $\dfrac12\implies a=?,b=?$
What about the other two?
