How do you simplify #log_7 9x+log_7x-3log_7 x#?

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Answer 1 · 2015-12-12T11:23:35

#log_7(9/x)#

First of all, observe that #3log_7(x)=log_7(x^3)#. So, your expression becomes

#log_7(9x)+log_7(x)-log_7(x^3)#

Now, the sum of two logarithms is the logarithm of the product:

#log_7(a)+log_7(b)=log_7(ab)#

So,

#color(green)(log_7(9x)+log_7(x))-log_7(x^3)=color(green)(log_7(9x^2))-log_7(x^3)#

And the difference of two logarithms is the logarithm of the ratio:

#log_7(a)-log_7(b)=log_7(a/b)#

So,

#log_7(9x^2)-log_7(x^3) = log_7(9x^2/x^3) = log_7(9/x)#