Question

How to solve this logarithmic function?

Answer 1

`x=1,2`

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(A)-log(B)=log(A//B)`:

`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 `2x-2=2(x-1)`. The relevance of this is that we can rewrite the equation as follows:

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

Now, we have a quadratic equation, if we let `t=3^(x-1)`.

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

And solve as normal:

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

`t=1,3`

Now we have to solve for `x` using `t=3^(x-1)`. In the case `t=1`:

`3^(x-1)=1`

Take the logarithm:

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

`x=1`

We also need the case `t=3`:

`3^(x-1)=3`

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

`x=2`