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SammyS

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log(b...We can represent log(a^b^c) as b^c log a. But what if b^c is still huge. Can we continue and end up with something like c* log(b) * log(a)? A quick test shows that this is probably not going to work. Any ideas? Thanks

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Ah, right, thank you. So do you think the general approach is otherwise correct?log(b^{c}⋅log(a)) = c⋅log(b)+log(log(a)) , not what you have.

Let's say in c⋅log(b)+log(log(a)) c is not just c but to the power of d. I am not sure, would I need to apply log to the sum? In other words, is there a general approach to get rid of all the nested exponentials by applying log consecutively? I am not sure what to do here:

log( c

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SammyS

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Once you have that sum, taking the log doesn't help.Ah, right, thank you. So do you think the general approach is otherwise correct?

Let's say in c⋅log(b)+log(log(a)) c is not just c but to the power of d. I am not sure, would I need to apply log to the sum? In other words, is there a general approach to get rid of all the nested exponentials by applying log consecutively? I am not sure what to do here:

log( c^{d}⋅log(b)+log(log(a)))

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