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

2 Answers
Ben
Mar 16, 2018

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)

Hp
Mar 16, 2018

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

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