Question

How do you solve `log_2(4x)=log_2(x+15)`?

Answer 1

Since both sides are taking the same logarithm, you can directly compare the terms inside the parentheses and find that `x=5`

Explanation:

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:

`log_2(4x)=n rArr 2^n=4x`

So if:

`2^n=4x`

and

`2^n=x+15`

then

`4x=x+15`

Now, we can solve for `x`:

`3x=15 rArr color(red)(x=5)`

Answer 2

`x=5`

Explanation:

`log_2(4x)=log_2(x+15)`

Or, `1= (log_2(x+15))/(log_2(4x))=log_(4x) (x+15)`

So,`(4x)^1=x+15`

Or, `3x=15`

So,`x=5`