How do you solve #log_5(2x+15)=log_5(3x)#?

1 Answer
Aug 3, 2016

Answer:

#x=15#

Explanation:

We start with #log_5(2x+15)=log_5(3x)#. In this case we got lucky and got an equation that has the same base, #5#. We can use this to our adantage, as long as we know how to convert a logarithm to an exponent. We do that using this rule: #log_color(red)(b)color(blue)(x)=color(green)(y)# becomes #color(red)(b)^color(green)(y)=color(blue)(x)#. The base stays the same and #x# and #y# switch.

Let's set up our conversion: #log_color(red)(5)color(blue)(2x+15)=color(green)(log_5(3x))# becomes #color(red)(5)^color(green)(log_5(3x))=color(blue)(2x+15)#. Now that we've got this set up, let's clean it up. Notice the #color(red)(5)^color(green)(log_5)# part. These two are inverses of eachother, so they cance out. That leaves us with #cancel(color(red)(5))^cancel(color(green)(log_5))color(white)(1)^color(green)(3x)=color(blue)(2x+15)#. The #3x# dops down and gives us #3x=2x+15#. That can be simplifed by subtracting #2x# on both sides, which leaves us with #x=15#.