If $a$, $b$ and $c$ are in geometric progression, then what are $\log_ax$, $\log_bx$ and $\log_cx$ in?
What I did: I substituted values for $x, a, b$ and $c$ and tried to solve it further.
What I got: Using the value of x as 2 and a = 2 b = 4 c = 8: I obtained that it is in harmonic progression. $$\log_22, \log_42, \log_82$$ $$1, \frac{1}{2}, \frac{1}{3}$$
However, if I used x = 64, then I am not getting a harmonic progression. Instead, I got: $$\log_264, \log_464, \log_864$$ $$6, 3, 2$$ I don't think that these numbers are in harmonic progression. What did I do wrong?
Btw. I know the answer is harmonic progression
Hint: $$\log_y(x)=\frac{\log(x)}{\log(y)}$$ Now try examining $$\frac{1}{\log_b(x)}-\frac{1}{\log_a(x)}$$ and $$\frac{1}{\log_c(x)}-\frac{1}{\log_b(x)}$$
Keep in mind that $\log(u)-\log(v)=\log(\frac{u}{v})$.