Question

Let $a>0,c>0,b=\sqrt{ac},a,c$ and $ac\ne 1,N>0$, then which of the following is true?

• A. $\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N}$
• B. $\frac{{\log }_{c}N}{{\log }_{a}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N}$
• C. $\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{b}N-{\log }_{a}N}{{\log }_{b}N-{\log }_{c}N}$
• D. $\frac{{\log }_{c}N}{{\log }_{a}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{c}N-{\log }_{b}N}$

Answer 1

The correct option is A $\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N}$

Let ${\log }_{a}N=x,{\log }_{c}N=y,{\log }_{\sqrt{ac}}N=z$
From last relation
${\log }_{\sqrt{ac}}\left(N\right)=z$
$N=(\sqrt{ac}{)}^z=(ac{)}^{\frac{z}{2}}$
$\Rightarrow {\log }_{a}N=\frac{z}{2}(1+{\log }_{a}c)\Rightarrow x=\frac{z}{2}(1+{\log }_{a}c)$
& ${\log }_{c}N=\frac{z}{2}(1+{\log }_{c}a)\Rightarrow y=\frac{z}{2}(1+{\log }_{c}a)$
or $\frac{2x}{z}-1={\log }_{a}c$ & $\frac{2y}{z}-1={\log }_{c}a$
$(\frac{2x}{z}-1)(\frac{2y}{z}-1)=1$
$\frac{4xy}{z^2}-\frac{2x}{z}-\frac{2y}{z}=0$
$\Rightarrow \frac{2xy}{z}-x-y=0$
$\Rightarrow 2xy-zx-yz=0$
$\Rightarrow (xy-yz)=zx-xy$
$\Rightarrow \frac{x}{y}=\frac{x-z}{z-y}or\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N}$