$$3^{\log(x)}-2^{\log(x)}=2^{\log(x+1)}-3^{\log(x-1)}$$ thanks for the correction Michael rozenberg. Now if you could please tell me how to solve this problem, I would appreciate it.
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See here for writing tips. – Sep 30 '17 at 06:54
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Put the exponent in {} so 2^log(x+1) becomes 2^{log(x+1)}. (The {} say to treat everything in as a chunk. If you leave it out only the first letter is formatted.) Put a \ before log to indicate it is a special math function so 2^{\log (x+1)}. The put everthing between $ signs: So $2^{\log (x+1)}$ which will be rendered $2^{\log (x+1)}$. – fleablood Sep 30 '17 at 06:59
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It's $$3^{\log(x)}-2^{\log(x)}=2^{\log(x+1)}-3^{\log(x-1)},$$ for which you can write the following:
3^{\log(x)}-2^{\log(x)}=2^{\log(x+1)}-3^{\log(x-1)}
Michael Rozenberg
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