Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$

Let $a,b$, and $c$ be positive real numbers such that

$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$

and

$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$

What is the value of

$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1) ?$$

I tried to convert the entire thing to fractional logs and multiply the expression and add the two equations but it did not help.

Solution 1:

As lab bhattacharjee commented, convert to natural logarithms and get $$\log_{a}b + \log_{b}c + \log_{c}a = 8\implies \frac{\log (b)}{\log (a)}+\frac{\log (a)}{\log (c)}+\frac{\log (c)}{\log (b)}=8$$ $$\log_{b}a + \log_{c}b + \log_{a}c = 13\implies \frac{\log (a)}{\log (b)}+\frac{\log (c)}{\log (a)}+\frac{\log (b)}{\log (c)}=13$$ Now expand $$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1)$$ which is $$\frac{\log (a)}{\log (b)}+\frac{\log (b)}{\log (a)}+\frac{\log (a)}{\log (c)}+\frac{\log (c)}{\log (a)}+\frac{\log (c)}{\log (b)}+\frac{\log (b)}{\log (c)}+2$$ I am sure that you can take it from here and finish.

Solution 2:

Since I prefer working with exponents rather than logarithms, here is how you could tackle it that way, using your quite possibly better intuition at handling powers:

$$x + y + z = 8$$

$$u + v + w = 13$$

With:

$$ a^x = b\,,\, b^y = c\,,\, c^z = a $$ and $$ b^u = a\,,\, c^v = b\,,\, a^w = c $$

From this you can see that:

$$ a^{x\,u} = b^u = a \; \Rightarrow \; x u = 1$$

So the second equation is actually:

$$ 1/x + 1/y + 1/z = 13 $$

If you add the fractions:

$$ yz + xz + xy = 13 \, x\,y\,z $$

Since: $$ (1+x)(1+y)(1+z) = xyz + xz + yz + xy + x + y + z + 1 $$

We need to figure out what is $xyz$. But:

$$ a^{x\,y\,z} = b ^{y\,z} = c^z = a \; \Rightarrow \; x y z = 1 $$

Now we have all the information we need and the result is $23$.

Solution 3:

We have the following:

$$(log_ab+1)(log_bc+1)(log_ca+1)=(\frac {logb}{loga}+1)(\frac {logc}{logb}+1)(\frac {loga}{logc}+1)$$

Expanding yields:

$$(\frac {logc}{loga}+\frac {logb}{loga}+\frac {logc}{logb}+1)(\frac {loga}{logc}+1)$$

$$=(\frac {loga}{loga}+\frac {logc}{loga}+\frac {logb}{logc}+\frac {logb}{loga} +\frac {loga}{logb}+\frac {logc}{logb}+\frac {loga}{logc}+1)$$

$$=(log_aa+log_ac+log_cb+log_ab+log_ba+log_bc+log_ca+1)$$

$$=(log_ab+log_bc+log_ca)+(log_ba+log_cb+log_ac)+2$$

Thus finally we see, via your initial condition:

$$(log_ab+1)(log_bc+1)(log_ca+1)=8+13+2=23$$

Solution 4:

Notice that $$ \log_uv\log_vw=\dfrac{\log v}{\log u}\cdot\dfrac{\log w}{\log v}=\dfrac{\log w}{\log u}=\log_uw \quad \forall u,v,w>0 $$ therefore \begin{eqnarray} (\log_ab+1)(\log_bc+1)(\log_ca+1)&=&(\log_ab\log_bc+\log_ab+\log_bc+1)(\log_ca+1)\\ &=&(\log_ac+\log_ab+\log_bc+1)(\log_ca+1)\\ &=&\log_ac\log_ca+\log_ac+\log_ab\log_ca+\log_ab+\log_bc\log_ca+\log_bc+\log_ca+1\\ &=&\log_aa+\log_ac+\log_cb+\log_ab+\log_ba+\log_bc+\log_ca+1\\ &=&(1+\log_aa)+(\log_ab+\log_bc+\log_ca)+(\log_ba+\log_cb+\log_ac)\\ &=&1+1+8+13=23. \end{eqnarray}