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.
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.
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 $23$.
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:
$$(log_ab+1)(log_bc+1)(log_ca+1)=8+13+2=23$$
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}
Here's how I solved this:
First, expand the expression. For conciseness, express it as $(x+1)(y+1)(z+1)$:
$$(x+1)(y+1)(z+1) = xyz + \Big(xy + yz + xz\Big) + \Big(x + y + z\Big) + 1$$
We recognize immediately that $x+y+z$ is simply the first equation we're given, with value 8.
It must be the case that the remaining terms are related (at least partially) to the second equation, so let's play with that a little bit. Is there a relationship between the second equation and the objects in the first?
If you don't know the relationship off the top of your head (I had forgotten) $\log_b{a}$ certainly looks close to $\log_a{b}$; expressing
$$\log_b{a} = x$$
$$\Leftrightarrow b = a^x$$
$$\Leftrightarrow 1 = x \log_b{a}$$
We quickly see that $\log_b{a} = \frac1{\log_a{b}} = \frac1x$, so the second equation is just
$$\frac1x + \frac1y + \frac1z = 13$$
The natural thing to do is to combine the terms on the LHS to get
$$\frac{xy + xz + yz}{xyz} = 13$$
This is excellent, since it directly relates to the unaccounted terms from our above expression and tells us that $xy+xz+yz=13xyz$, so
$$(x+1)(y+1)(z+1) = 14xyz + 9$$
So we're left with needing to know the value for $xyz = \log_a{b} \log_b{c} \log_c{a}$, which looks cyclical. Recalling that $\log_b{a}$ and $\log_a{b}$ have an inverse/cyclical relationship, that $\log_b{a} \log_a{b} = 1$, we see if a similar identity might hold for this extended case:
$$\log_a{b} \log_b{c} \log_c{a} = 1$$
We can quickly verify this is indeed the case, so that $xyz = 1$ and the result that $(x+1)(y+1)(z+1) = 23$ follows immediately.
