On logarithm equations

After posting this problem I used some of my time on solutions to equations of logarithms and I found the result as follows:

The system of equation (given below) $\log_{b_1}w=a_1$ $\log_{b_2}w=a_2$ $\log_{b_3}w=a_3$ $.$ $.$ $.$ $\log_{b_n}w=a_n$ $\log_{b_1b_2b_3...b_{n-1}b_nb_{n+1}}w=a_{n+1}$ $Then,$ $\boxed{\log_{b_{n+1}}w=\frac{a_{n+1}}{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}}$ This is only if $\boxed{w≠0}$

Proof: From the system of equation (given below) $\log_{b_1}w=a_1$ $\log_{b_2}w=a_2$ $\log_{b_3}w=a_3$ $.$ $.$ $.$ $\log_{b_n}w=a_n$ $\log_{b_1b_2b_3...b_{n-1}b_nb_{n+1}}w=a_{n+1}$ We get $b_1^{a_1}=w$ $b_2^{a_2}=w$ $b_3^{a_3}=w$ $.$ $.$ $.$ $b_n^{a_n}=w$ $(b_1b_2b_3...b_nb_{n+1})^{a_{n+1}}=w$ Simplifying above expression $b_1^{a_{n+1}}b_2^{a_{n+1}}b_3^{a_{n+1}}...b_n^{a_{n+1}}b_{n+1}^{a_{n+1}}=w$ $(b_1^{a_1})^{\frac{a_{n+1}}{a_1}}(b_2^{a_2})^{\frac{a_{n+1}}{a_2}}(b_3^{a_3})^\frac{a_{n+1}}{a_3}...(b_n^{a_n})^\frac{a_{n+1}}{a_n}(b_{n+1})^{a_{n+1}}=w$ $(w)^\frac{a_{n+1}}{a_1}(w)^\frac{a_{n+1}}{a_2}(w)^\frac{a_{n+1}}{a_3}...(w)^\frac{a_{n+1}}{a_n}(b_{n+1})^{a_{n+1}}=w$ $w^{\frac{a_{n+1}}{a_1}+\frac{a_{n+1}}{a_2}+\frac{a_{n+1}}{a_3}...\frac{a_{n+1}}{a_n}}b_{n+1}^{a_{n+1}}=w$ $w^{a_{n+1}(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}...\frac{1}{a_n})}b_{n+1}^{a_{n+1}}=w$ $w^{a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}b_{n+1}^{a_{n+1}}=w$ $b_{n+1}^{a_{n+1}}=\frac{w}{w^{a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}}$ $b_{n+1}^{a_{n+1}}=w^{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}$ Taking power $\frac{1}{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}$ on both sides $b_{n+1}^{\frac{a_{n+1}}{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}}=w^{\frac{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}}=w$ Taking $\log_{b_{n+1}}$ on both side $\boxed{\log_{b_{n+1}}w=\frac{a_{n+1}}{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}}$