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

2 Answers
Mar 16, 2018

Answer:

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)#

Mar 16, 2018

Answer:

#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#