How do you solve #log_25(x+9) = 3/2#?

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Answer 1 · 2016-02-04T00:24:13

#x=116#

Using the property that #a^(log_a(x)) = x#, we have

#log_25(x+9)=3/2#

#=> 25^(log_25(x+9)) = 25^(3/2)#

#=>x+9 = 125#

#:. x = 116#

Answer 2 · 2016-02-04T00:24:56

I found: #x=116#

We can use the definition of log:
#log_ax=y ->x=a^y#
and get:
#x+9=25^(3/2)#
#x+9=sqrt(25^3)#
where we used the fact that: #x^(a/b)=rootbx^a#

#x+9=25*5#

#x+9=125#

#x=116#