If log644 = x, then 64x = 4. The cubed root of 64 (which is the same as 641/3) is 4, so log base 64 of 4 is 1/3.
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log 100 base e = log 100 base 10 / log e base 10 log 100 base 10 = 10g 10^2 base 10 = 2 log 10 base 10 = 2 log e base 10 = 0.434294 (calculator) log 100 base e = 2/0.434294 = 4.605175
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
k=log4 91.8 4^k=91.8 -- b/c of log rules-- log 4^k=log 91.8 -- b/c of log rules-- k*log 4=log91.8 --> divide by log 4 k=log 91.8/log 4 k= 3.260
You can convert to the same base, by the identity: logab = log b / log a (where the latter two logs are in any base, but both in the same base).
log2x = log x / log 2 On the right side, you can use logarithm in any base (calculators usually provide base-10 and base-e), just be sure to use the same base in both cases. Thus: log2x = ln x / ln 2 or: log2x = log10x / log102