If $a=1+\log_xyz, b=1+\log_yzx, c=1+\log_zyx$, then $ab+bc+ac=$

Question

MY SOLUTION

Let’s take a. So $1=\log_xx$ Therefore $$a=\log_xxyz$$ Similarly $$b=\log_yxyz$$ and $$c=\log_zxyz$$ In the original question, the expression becomes $$(\log_xxyz)(\log_yxyz)+(\log_yxyz)(\log_zxyz)+(\log_zxyz)(\log_xxyz)$$ I could not solve further. I tried all sorts of tricks and properties, but I can’t arrive at the answer. How should I proceed?

Answer is abc

Answer 1

Note that $a = \log_x{xyz} = \frac{\ln(xyz)}{\ln(x)}$ And similarly with $b,c$, so we get:

$ab + bc + ac = \ln^2(xyz)(\frac{1}{\ln(x)\ln(y)} + \frac{1}{\ln(x)\ln(z)} + \frac{1}{\ln(y)\ln(z)}) = \ln^2(xyz)(\frac{\ln(z)+\ln(y)+\ln(x)}{\ln(x)\ln(y)\ln(z)}) = \frac{\ln^3(xyz)}{\ln(x)\ln(y)\ln(z)} = \frac{\ln(xyz)}{\ln(x)} \cdot \frac{\ln(xyz)}{\ln(y)} \cdot \frac{\ln(xyz)}{\ln(z)} =abc$

Answer 2

$$a=\log_xx+\log_x(yz)=\log_x(xyz)$$

$$\dfrac1a=\log_{xyz}x$$

$$\dfrac1a+\dfrac1b+\dfrac1c=\cdots=\log_{xyz}(xyz)=1$$