How to convert log(base7) of 2x-1 to log that has the base of 5?
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First, I have this formula memorized
log_7 (2x-1) = log_b(2x-1) / log_b(7) where b can be any base. I usually use natural logarithms (base e) but it could be base 10 or base 2 .... or base 5.
log_7(2x-1) = log_5(2x-1) / log_5(7)
log_5(2x-1) = log_5(7) log_7(2x-1) So that might be your answer already.
But if you want that to be in terms of some more usual base, then use the same formula again for converting the base 5 log and the base 7 log to a general base.
log_5(2x-1) = ( log_b(7) / log_b(5) ) (log_b(2x-1) / log_b(7) )
The log_b(7) cancel out so what you have left is just log_b(2x-1) / log_b(5), basically showing the formula is consistent.
log_7 (2x-1) = log_b(2x-1) / log_b(7) where b can be any base. I usually use natural logarithms (base e) but it could be base 10 or base 2 .... or base 5.
log_7(2x-1) = log_5(2x-1) / log_5(7)
log_5(2x-1) = log_5(7) log_7(2x-1) So that might be your answer already.
But if you want that to be in terms of some more usual base, then use the same formula again for converting the base 5 log and the base 7 log to a general base.
log_5(2x-1) = ( log_b(7) / log_b(5) ) (log_b(2x-1) / log_b(7) )
The log_b(7) cancel out so what you have left is just log_b(2x-1) / log_b(5), basically showing the formula is consistent.
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log [base 7] (2x - 1) = log [base 5] (2x - 1) / log [base 5] 7
Yes, log [5] 7 will be irrational. Give the above as the exact answer or use calculator approximation with as many decimal places as you think right.
Yes, log [5] 7 will be irrational. Give the above as the exact answer or use calculator approximation with as many decimal places as you think right.