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) logb1w=a1\log_{b_1}w=a_1 logb2w=a2\log_{b_2}w=a_2 logb3w=a3\log_{b_3}w=a_3 .. .. .. logbnw=an\log_{b_n}w=a_n logb1b2b3...bn1bnbn+1w=an+1\log_{b_1b_2b_3...b_{n-1}b_nb_{n+1}}w=a_{n+1} Then,Then, logbn+1w=an+11an+1s=1n1as\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 w0\boxed{w≠0}

Proof: From the system of equation (given below) logb1w=a1\log_{b_1}w=a_1 logb2w=a2\log_{b_2}w=a_2 logb3w=a3\log_{b_3}w=a_3 .. .. .. logbnw=an\log_{b_n}w=a_n logb1b2b3...bn1bnbn+1w=an+1\log_{b_1b_2b_3...b_{n-1}b_nb_{n+1}}w=a_{n+1} We get b1a1=wb_1^{a_1}=w b2a2=wb_2^{a_2}=w b3a3=wb_3^{a_3}=w .. .. .. bnan=wb_n^{a_n}=w (b1b2b3...bnbn+1)an+1=w(b_1b_2b_3...b_nb_{n+1})^{a_{n+1}}=w Simplifying above expression b1an+1b2an+1b3an+1...bnan+1bn+1an+1=wb_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 (b1a1)an+1a1(b2a2)an+1a2(b3a3)an+1a3...(bnan)an+1an(bn+1)an+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)an+1a1(w)an+1a2(w)an+1a3...(w)an+1an(bn+1)an+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 wan+1a1+an+1a2+an+1a3...an+1anbn+1an+1=ww^{\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 wan+1(1a1+1a2+1a3...1an)bn+1an+1=ww^{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 wan+1s=1n1asbn+1an+1=ww^{a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}b_{n+1}^{a_{n+1}}=w bn+1an+1=wwan+1s=1n1asb_{n+1}^{a_{n+1}}=\frac{w}{w^{a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}} bn+1an+1=w1an+1s=1n1asb_{n+1}^{a_{n+1}}=w^{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}} Taking power 11an+1s=1n1as\frac{1}{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}} on both sides bn+1an+11an+1s=1n1as=w1an+1s=1n1as1an+1s=1n1as=wb_{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 logbn+1\log_{b_{n+1}} on both side logbn+1w=an+11an+1s=1n1as\boxed{\log_{b_{n+1}}w=\frac{a_{n+1}}{1-a_{n+1}\sum_{s=1}^{n}\frac{1}{a_s}}}

Note by Zakir Husain
4 weeks ago

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@Zakir Husain, please post more PRMO problems. Thank you!

Mahdi Raza - 4 weeks ago

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Got to agree with you!

Neil Chaturvedi - 4 weeks ago

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Neat substitutions! Nice!

Mahdi Raza - 4 weeks ago

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