How do you solve #log_2(4x)=log_2(x+15)#?
Since both sides are taking the same logarithm, you can directly compare the terms inside the parentheses and find that
The easy way to look at this is by realizing that you can compare the in-parentheses terms directly. This is because they're both logarithms of the same base.
A base-2 log has equivalence like this:
Now, we can solve for