How do you solve #2log_5x = log_5 9#?

1 Answer
Feb 24, 2016

First simplify using the rule #alogn = logn^a#

Explanation:

#log_5(x^2) = log_5(9)#

Put everything to one side of the equation.

#0 = log_5(9) - log_5(x^2)#

Simplify using the rule #log_am - log_an = log_a(m / n)#

#0 = log_5(9/x^2)#

Convert to exponential form.

#5^0 = 9/x^2#

#1 = 9/x^2#

#x^2 = 9#

#x = +-3#

However, only +3 works as a solution because the log of a negative number is non-defined.

The solution set is x = 3.

Hopefully this helps!