How do you condense #6log(x)-log(17)# to a simple logarithm?

1 Answer
Matthew Liu
Jun 18, 2018

#log(x^6/17)#

Explanation:

There are some laws of logarithms. One of them is the powers of logarithms. For instance, if we have the function #f(x)= log_a(b^c)# , then we can rewrite this as #f(x)=clog_a(b)#. Since we are trying to condense the expression into a single logarithm, #6log(x)# is equal to #log(x^6)#. Another law of logarithms is if we have the expression #log_a(b)-log_a(c)#, we can rewrite this as #log_a(b/c)#. So #log(x^6)-log(17)=log(x^6/17)#

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