How do you solve log_25(x+9) = 3/2log25(x+9)=32?

2 Answers
Feb 4, 2016

x=116x=116

Explanation:

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

log_25(x+9)=3/2log25(x+9)=32

=> 25^(log_25(x+9)) = 25^(3/2)25log25(x+9)=2532

=>x+9 = 125x+9=125

:. x = 116

Feb 4, 2016

I found: x=116

Explanation:

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