How do you solve #(18) ^ { - 3} ( 18) ^ { n } = 18^ { - 11}#?

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Answer 1 · 2018-01-16T02:36:09

#n=-8#

First, remember that: #a^b*a^c=a^(b+c)#

Therefore, #18^-3*18^n=18^(-3+n)#

Now, since both #18^(-3+n)# and #18^-11# have the base of 18, we can say that #-3+n=-11#.

Therefore,
#n=-8#

Answer 2 · 2018-08-06T03:16:59

#n=-8#

Recall the exponent property

#x^ax^b=x^(a+b)#

When we multiply exponents, we add the powers. To find #n#, we can essentially setup the equation

#-3+n=-11#

Adding #3# to both sides, we get

#n=-8#

Hope this helps!