How do you evaluate # ln (2x+3)=7#?

1 Answer
Simranjit D.
Mar 10, 2018

#x~~546.817# or #x=(e^7 -3)/2#

Explanation:

Remember, anytime you see ln, it is always equal to
#loge# (the e is a subscript). You work with them the same way as you would any other log.
Problem:
#ln(2x+3)=7#
Step 1: take #loge# of both sides.
#ln(2x+3)=ln(e^7)#
Step 2: Since the logs have the same bases, make the #(2x+3)=(e^7)# equal to each other
Step 3: Solve so the x is on the left side by itself:
#x=(e^7 -3)/2#
Step 4: if you need the decimal form, just plug in 2.718281828 (Euler's number) for e and solve.
#x=(2.718281828 ^7 -3)/2 => x~~546.817#

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