Interesting logarithms - 2

We can convert all of the logarithms into an exponential form:

$a^{15}=w$ $b^{12}=w$ $c^{20}=w$ $d^{60}=w$ $a^{\frac{60}{13}}b^{\frac{60}{13}}c^{\frac{60}{13}}d^{\frac{60}{13}}x^{\frac{60}{13}}=w$

We can also change the last one so that we can substitute $w$ :

$a^{\frac{60}{13}}b^{\frac{60}{13}}c^{\frac{60}{13}}d^{\frac{60}{13}}x^{\frac{60}{13}}=w \Longrightarrow \left(a^{15}\right)^{\frac{4}{13}} \left(b^{12}\right)^{\frac{5}{13}} \left(c^{20}\right)^{\frac{3}{13}} \left(d^{60}\right)^{\frac{1}{13}}x^{\frac{60}{13}}=w$

We can substitute $w$ in now and it cleans itself up:

$\left(a^{15}\right)^{\frac{4}{13}} \left(b^{12}\right)^{\frac{5}{13}} \left(c^{20}\right)^{\frac{3}{13}} \left(d^{60}\right)^{\frac{1}{13}}x^{\frac{60}{13}}=w \Longrightarrow w^{\frac{4}{13}} w^{\frac{5}{13}} w^{\frac{3}{13}} w^{\frac{1}{13}}x^{\frac{60}{13}}=w \Longrightarrow wx^{\frac{60}{13}}=w$

Once we solve for x we get:

$wx^{\frac{60}{13}}=w \Longrightarrow x^{\frac{60}{13}}=1 \Longrightarrow x=1$

Using this knowledge, we see that $log_xw$ is undefined because taking $log_1n, n\neq1$ is impossible and if $n=1$ there is infinite answers

Applying this formula $\log_xw=\frac{1}{1-\frac{60}{13}\times(\frac{1}{15}+\frac{1}{12}+\frac{1}{20}+\frac{1}{60})}=\frac{1}{1-\frac{60}{13}\times(\frac{13}{60})}=\frac{1}{0}$ The value we find is undefined hence the answer is not possible