How do you solve #ln(4x+1)=ln(2x+5)#?

1 Answer
Jul 11, 2016

#x = 2#

Explanation:

Remembering that taking the exponential of a natural logarithm gets rid of the natural logarithm, we can apply it in this case to have a simpler equation, giving us

#cancel(e)^(cancel(ln)(4x+1)) = cancel(e)^(cancel(ln)(2x+5))#

#4x+1 = 2x+5#

Now, subtracting #2x# from both sides yields

#2x+1 = 5#

Subtracting #1# from both sides gives us

#2x = 4#, so #x = 2#