How do you solve #Log_(5)x + log_(3)x = 1#?

1 Answer
Jun 27, 2016

Answer:

#x = 3^{(log_e 5)/(log_e 15)}#

Explanation:

We know that

#log_a b = (log_eb)/(log_e a)#

Then

#Log_(5)x + log_(3)x = 1->(log_e x)/(log_e 5)+(log_e x)/(log_e 3)=1#

or

#log_e x= (log_e 5 cdot log_e 3)/(log_e 5 + log_e 3)#

or

#x = e^{(log_e 5 cdot log_e 3)/(log_e 5 + log_e 3)}#

#x = 3^{(log_e 5)/(log_e 15)}#