# How to solve this logarithmic function?

##### 1 Answer

#### Explanation:

First, I would get both the logarithms on the same side.

#log_2(3^(2x-2)+7)-log_2(3^(x-1)+1)=2#

Now, use the rule

#log_2((3^(2x-2)+7)/(3^(x-1)+1))=2#

Rewrite by undoing the logarithm:

#(3^(2x-2)+7)/(3^(x-1)+1)=2^2=4#

Cross-multiply:

#3^(2x-2)+7=4(3^(x-1)+1)#

#3^(2x-2)+7=4(3^(x-1))+4#

#3^(2x-2)+3=4(3^(x-1))#

I stared at this for a little while before I realized something cool: that

#(3^(x-1))^2-4(3^(x-1))+3=0#

Now, we have a quadratic equation, if we let

#t^2-4t+3=0#

And solve as normal:

#(t-1)(t-3)=0#

#t=1,3#

Now we have to solve for

#3^(x-1)=1#

Take the logarithm:

#x-1=log_3(1)=0#

#x=1#

We also need the case

#3^(x-1)=3#

#x-1=log_3(3)=1#

#x=2#