if $x>1$ and $\log_2x,\log_3x,\log_x16$ are in G.P then what is x $=$

Question

if $x>1$ and $\log_2x,\log_3x,\log_x16$ are in Geometric progression then what is x equal to?

Solution:
$(\log_3x)^2=\log_2x\times\log_x16=\log_216=\log_22^4=4$
$\log_3x=2 or x=3^2=9$

so my doubt is how $(\log_3x)^2=\log_2x\times\log_x16$ while the rest i have understood

Answer 1

The equality comes from the geometric progression: $$\log_2 x = a, \quad \log_3 x = ar, \quad \log_x 16 = ar^2$$ Thus $$(\log_3 x)^2 = a^2 r^2 = a \cdot ar^2 = \log_2 x \log_x 16$$

Answer 2

I do not know if this is the kind of solution you expect, so forgive me if I am off-topic.

I think that things are easy if you go to natural logarithms or just to a common base; doing so and using ALexR notations, you have $$a=\frac{\log (x)}{\log (2)}$$ $$ar=\frac{\log (x)}{\log (3)}$$ $$ar^2=\frac{\log (16)}{\log (x)}$$ So, using $a \cdot ar^2=a^2r^2=(ar)^2$ $$\frac{\log (x)}{\log (2)}\times \frac{\log (16)}{\log (x)}=\Big(\frac{\log (x)}{\log (3)}\Big)^2$$ I am sure that you can take from here.

Answer 3

Remember that a G.P. is characterized by the fact that after the first one, any term in the progression equals the one that preceedes it multiplied by a constant. Thus, a G.P. looks like

$$a,\,aq,\,aq^2,\,...,\,aq^n,\,aq^{n+1},\ldots$$

Now, it's easy to check that if we take any three successive terms $\;a_{n-1},\,a_n,\,a_{n+1}\;$ in a G.P. we have that $\;a_n^2=a_{n-1}a_{n+1}\;$ .

Thus, in your case, as we're given those three successive (this is understood without saying, I think) elements in a G.P., you obtain

$$\log^2_3x=\log_2x\cdot\log_x16=4\log_2x\log_x2=4\implies\log_3x=\pm2\implies x=3^{\pm2}=\begin{cases}9\\{}\\\frac19\end{cases}$$