How do you solve #\log _ { 9} ( 2x + 5) = \log _ { 9} 5x#?

Question

1130 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-02T19:04:32

#x=5/3#

Because you have a #log_9# on both sides, raise both sides as powers of #9#, which effectively cancels out the #log#s.

#log_9(2x+5) = log_9(5x)#

#9^(log_9(2x+5)) = 9^(log_9(5x))#

#2x+5 = 5x#

Now you can simply solve

#2x+5=5x#

#5=3x#

#x=5/3#