How do you solve #2+log_3(2x+5)-log_3x=4#?

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Answer 1 · 2015-02-08T01:29:13

I would start by collecting the logs on one side:

#log_3(2x+5)-log_3(x)=4-2#

I can use the fact that:
#logM+logN=log(M/N)#
Giving:

#log_3((2x+5)/x)=2#

Use the definition of logarithm:

#log_ax=b ->a^b=x#

Giving:

#(2x+5)/x=3^2#
#2x+5=9x#
#7x=5#
#x=5/7#

hope it helps